We claim that the solutions are $a\:\diamondsuit\: b = a\cdot b$ and $a\:\diamondsuit\: b = \frac{a}{b}$. It's easy to see that these work; we now show that these are the only ones. Let $P(a,b,c)$ denote the given assertion. We first show that $\diamondsuit$ is "injective." Suppose that $a\:\diamondsuit\: b = a\:\diamondsuit\: c$ for some $a,b,c\:$; then by $P(a,a,b)$ and $P(a,a,c)$, we have $$(a\:\diamondsuit\: a)\cdot b = a\:\diamondsuit\: (a\:\diamondsuit\: b)=a\:\diamondsuit\: (a\:\diamondsuit\: c)=(a\:\diamondsuit\: a)\cdot c$$which implies that $b=c$. By $P(a,a,1)$ and the injectivity we just proved, we have $$a\:\diamondsuit\: (a\:\diamondsuit\: 1)=(a\:\diamondsuit\: a)\cdot1=a\:\diamondsuit\:a\implies a\:\diamondsuit\: 1=a.$$ By $P(a,1,c)$, we see that $$a\:\diamondsuit\: (1\:\diamondsuit\: c)=(a\:\diamondsuit\: 1)\cdot c = ac$$and in particular $$1\:\diamondsuit\: (1\:\diamondsuit\: c) = c.$$Now $P(a,1,1\:\diamondsuit\: c)$ gives $$a(1\:\diamondsuit\: c)=a\:\diamondsuit\: (1\:\diamondsuit\: (1\:\diamondsuit\: c)) = a\:\diamondsuit\: c.$$ From $P(1,b,1\:\diamondsuit\: c)$, we have $$(1\:\diamondsuit\: b)\cdot (1\:\diamondsuit\: c) = (1\:\diamondsuit\: (b\:\diamondsuit\: (1\:\diamondsuit\: c))))=1\:\diamondsuit\: (bc).$$Thus, the function $f(x)=(1\:\diamondsuit\: x)$ is multiplicative; making the transformation $$g(x)=\log (f(e^x))$$produces an additive $g:\mathbb{R}\to\mathbb{R}$. But we know that $$(1\:\diamondsuit\: a)=\frac{1}{a}\cdot (a\:\diamondsuit\: a)\geq \frac{1}{a}$$for all $a\geq 1$. Choosing $a\in [1,e]$ and thus $\log(a)\in [0,1]$ yields $$g(\log(a))=\log(f(a))\geq\log(\frac{1}{a})\geq -1$$so $g$ is bounded below on a nontrivial interval. Since $g$ is also additive, we conclude that $g(x)=kx$ for some $k\in\mathbb{R}$. Thus $f(x)=x^k$ and $$a\:\diamondsuit\: b = a(1\:\diamondsuit\: b) = af(b)=ab^k.$$Substituting this back into $P(a,b,c)$ gives $$a(bc^k)^k=ab^kc\implies k^2=1 \implies k=\pm 1$$so $a\:\diamondsuit\: b = a\cdot b$ or $a\:\diamondsuit\: b = \frac{a}{b}$.

The answer is normal multiplication $a\ \diamondsuit\ b = ab$ and normal division $a\ \diamondsuit\ b = a/b$. They clearly works. Let $P(a,b,c)$ denote the first given condition. First, $P(a,b,1)$ gives $a\ \diamondsuit\ (b\ \diamondsuit\ 1) = a\ \diamondsuit\ b$ thus replacing $P(a,b,c\ \diamondsuit\ 1)$ gives $a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot (c\ \diamondsuit\ 1)$. Therefore $c\ \diamondsuit\ 1 = c$. Now, $P(a,1,b)$ gives $a\ \diamondsuit (1\ \diamondsuit\ b) = ab$. Thus $P(a,b,1\ \diamondsuit\ c)$ gives $a\ \diamondsuit\ bc = (a\ \diamondsuit\ b)\cdot (1\ \diamondsuit\ c)$. In particular, this implies $1\ \diamondsuit\ xy = (1\ \diamondsuit\ x)(1\ \diamondsuit\ y)$ and $x\ \diamondsuit\ y = x\cdot(1\ \diamondsuit\ y)$ (by pluggin $a=1$ and $c=1$ respectively). This lead us to define a function $f(x) = (1\cdot x)$. Thus $f$ is multiplicative. Moreover $1\ \diamondsuit\ (1\ \diamondsuit\ x) = x$ (by $P(1,1,x)$) thus $f$ is an involution. Moreover, the second condition gives $f(x)\geqslant\tfrac{1}{x}$ for $x\geqslant 1$. It turns out that these are sufficient conditions to determine $a\ \diamondsuit\ b$. So, we focus on solving the new one-var FE. Let $g(x) = \ln f(e^x)$. Therefore $g(x+y)=g(x)+g(y)$ and $g(x)\geqslant -x$ for all $x\geqslant 0$. If $g$ is non-linear, then it's a well known theorem that the graph of any nonlinear Cauchy function is dense in a plane. However, it's clear that $f$ is bounded below in $[2019,10^{10}]$ thus it's contradiction. Hence $g(x)=cx$ for some constant $c$ which implies $f(x)=x^c$. From $f(f(x))=x$, we get $c=\pm 1$. This gives the two solutions described at the beginning.

Cool problem. Let $P(a,b,c)$ be the assertion $a\ast (b\ast c)=(a\ast b)\cdot c$. Note that, if for a fixed $b$, $b\ast c=b\ast d$, then comparing $P(a,b,c)$ and $P(a,b,d)$ we get $c=d$. Thus, for any fixed $a\in\mathbb{R}^+$, $x\mapsto a\ast x$ is injective. Next, $P(a,a,1)$ implies that, $a\ast (a\ast 1)=a\ast a$, and thus, $a\ast 1=a$, using the injectivity. Now, $P(1,1,a)$ implies $1\ast (1\ast a)=a$. As done above, set $f(x)=1\ast x$. This expression yields $f(f(x))=x$. Next, $P(a,1,c)$ yields $a\ast(1\ast c)=ac$, that is, $a\ast f(c)=ac$. In particular, inputting $c\mapsto f(c)$ here, and using $f(f(c))=c$, we get that $a\ast c = af(c)$. In particular, it suffices to determine $f(\cdot)$ to recover the operation $\ast$. Now, using this repeatedly, $a\ast (b\ast c)= a\ast(bf(c))=af(bf(c))$, which is equal to $(a\ast b)\cdot c=af(b)c$, that is, $f(bf(c))=f(b)c$ for every $b,c\in\mathbb{R}^+$. Note, plugging $c\mapsto f(c)$ here, we get $f(bf(f(c)))=f(bc)=f(b)f(c)$. Let $b=e^k,c=e^\ell$, this yields $f(e^{k+\ell})=f(e^k)f(e^\ell)$. Set $g(x)=f(e^x)$, this yields $g(k+\ell)=g(k)g(\ell)$. Finally, let $\varphi(x)=\log g(x)$, this brings us $\varphi(x+y)=\varphi(x)+\varphi(y)$. (I believe using second condition), we get $\varphi(x)=\alpha x$ for some $\alpha$ constant, which yields, $g(x)=f(e^x)=e^{\alpha x}$, and thus, $f(k)=k^\alpha$ for some $\alpha$ constant. Plugging this for $f(f(k))=k^{\alpha^2}=k$, we get $\alpha=\pm 1$, yielding $a\ast c=af(c)=a/c$ or $ac$, as claimed.

The answer is $\boxed{\times}$ and $\boxed{\div}$. It is easy to check that these work. We now show these are the only solutions. Claim 1. If $a\mathbin{\diamondsuit}p=a\mathbin{\diamondsuit}q$, then $p=q$. Proof. Check that $$(a\mathbin{\diamondsuit}a)\cdot p=a\mathbin{\diamondsuit}(a\mathbin{\diamondsuit}p)=a\mathbin{\diamondsuit}(a\mathbin{\diamondsuit}q)=(a\mathbin{\diamondsuit}a)\cdot q,$$whence $p=q$. $\blacksquare$ Claim 2. If $p\mathbin{\diamondsuit}a=q\mathbin{\diamondsuit}a$, then $p=q$. Proof. Check that $$(a\mathbin{\diamondsuit}p)\cdot a=a\mathbin{\diamondsuit}(p\mathbin{\diamondsuit}a)=a\mathbin{\diamondsuit}(q\mathbin{\diamondsuit}a)=(a\mathbin{\diamondsuit}q)\cdot a,$$whence $a\mathbin{\diamondsuit}p=a\mathbin{\diamondsuit}q$ and $p=q$. $\blacksquare$ Claim 3. $a\mathbin{\diamondsuit}1=a$ and $1\mathbin{\diamondsuit}(1\mathbin{\diamondsuit}a)=a$. Proof. For the first part, note that $a\mathbin{\diamondsuit}(a\mathbin{\diamondsuit}1)=(a\mathbin{\diamondsuit}a)\cdot 1=a\mathbin{\diamondsuit}a$, so by Claim 1, $a\mathbin{\diamondsuit}1=a$, as desired. For the second part, check that $1\mathbin{\diamondsuit}(1\mathbin{\diamondsuit}a)=(1\mathbin{\diamondsuit}1)\cdot a=a$. $\blacksquare$ Claim 4. $a\mathbin{\diamondsuit}b=a\cdot(1\mathbin{\diamondsuit}b)$. Proof. Check that $$a\mathbin{\diamondsuit}b=a\mathbin{\diamondsuit}\big(1\mathbin{\diamondsuit}(1\mathbin{\diamondsuit}b)\big)=(a\mathbin{\diamondsuit}1)\cdot(1\mathbin{\diamondsuit}b)=a\cdot(1\mathbin{\diamondsuit}b),$$as requested. $\blacksquare$ Now, let $f(b)=1\mathbin{\diamondsuit}b$, so that $a\mathbin{\diamondsuit}b=af(b)$. Our given functional equation rewrites to $$af\big(bf(c)\big)=acf(b)\implies f\big(bf(c)\big)=cf(b).$$However, by Claim 1, $f$ is injective, and by Claim 4, $f$ is an involution, so plugging in $f(c)$ as $c$ gives $f(bc)=f(b)f(c)$. Hence, $g(x)=\ln f(e^x)$ satisfies $g(b+c)=g(b)+g(c)$. The second condition implies that $g(x)\ge -x$ for all $x\ge 0$, so $g$ is bounded, and thus $g(x)=kx$ for some $k$. It follows that $f(x)=x^k$, but substituting yields $k=\pm 1$. Hence, $a\mathbin{\diamondsuit}b=a\cdot b$ or $a\div b$, as desired. $\square$

The good old multiplicative-to-Cauchy conversion! I have a somewhat different solution with a similar main idea. Let $a\,\diamondsuit\, b$ be denoted by $f(a, b)$. Then, it's standard to prove that: $f(a, x)$ takes all positive real values as we vary $x$ keeping $a$ fixed. $(1)$ $f(a, b) = f(a, c) \implies b = c$ $(2)$ $f(b, a) = f(c, a) \implies b = c$ $(3)$ $f(a, 1) = a$ $(4)$ Now, $f(a, b) = a*f(a, b)/a = f(a, 1)*f(a, b)/a = f(a, f(1, f(a, b)/a))$ So, $b = f(1, f(a, b)/a)$, so $f(a, b)/a$ is constant for fixed $b$. So, we can define a function $g(x)$ over the positive reals s.t. $f(a, b) = a*g(b)$ Note that $(1)$ implies that $g$ is surjective, and $(4)$ gives $g(1) = 1$. The first assertion given in the problem implies that $g(bg(c)) = cg(b)$. Putting $b = 1$ gives $g(g(c)) = c$. As $g$ is surjective, we have that $g$ is multiplicative. Define $h(x)$ over the real numbers by $h(x) = ln(g(e^x))$. Note that $g$ multiplicative gives $h$ additive. Note that the second assertion of the problem statement implies $x + h(x) \ge 0$ for all $x \ge 0$. So $h$ is bounded below in $(0, 1)$, so $h(x) \equiv ax$ for some real constant $a$. This gives $g(x) \equiv x^a$. As $g(g(x)) = x$, $a = 1$ or $a = -1$, so $f(a, b) \equiv a/b$ or $f(a, b) \equiv ab$. It's easy to see that both of them work, and we're done.

How rigorous is it to go directly from $h(bc)=h(b)h(c)$ to $h \equiv x^p$? for some power $p$, i.e. can this be stated as well known? Or would one need to reduce it to the lemma used in the official solution (pg 2) in order to get points? edit: yeah ok pathological functions are bad

mathisawesome2169 wrote: How rigorous is it to go directly from $h(bc)=h(b)h(c)$ to $h \equiv x^p$? for some power $p$, i.e. can this be stated as well known? It's not true By the way, for anyone that doesn't understand the title reference, compare AMC 2016 10A #23 / 12A #20 from which this problem came.

^Thanks to the AMC problem, I realized $a/b$ was a solution way before I found that $ab$ was a solution...

MathStudent2002 wrote: ^Thanks to the AMC problem, I realized $a/b$ was a solution way before I found that $ab$ was a solution... I realized that multiplication was a solution once I already had all the ideas to solve the problem. Maybe I should do less AMC grinding...

Here is my solution without using Cauchy.

I wonder how much easier/harder this problem would have been if the symbol was changed.

Not that much? I mean I just replaced the symbol with "x" although this did get a little confusing with multiplication... Maybe a carrot would have been better.

I just want to talk about my solution starting from letting $f(c)=1\diamond c$ and $f$ has already been found to be a multiplicative involution on $\mathbb{R}_{>0}$ with $f(1)=1$ and satisfying another condition which becomes $af(a)\geq1$ if $a\geq1$ and $af(a)\leq1$ if $a\leq1$. It suffices to show that $f=id$ or $f=\frac{1}{id}$ (the identity function or the reciprocal function). Let $\mathbb{R}_{>0}^{f=id}$ denote the fixed points of $f$. Consider the function $g: \mathbb{R}_{>0}\to \mathbb{R}_{>0}^{f=id}$ defined by $g(a)=a^{\frac{1}{2}}f(a^{\frac{1}{2}})$. It has the property that $g(\mathbb{R}_{>0}^{f=id})=\mathbb{R}_{>0}^{f=id}$. If it were injective (one to one) then $\mathbb{R}_{>0}^{f=id}=\mathbb{R}_{>0}$ and $f$ is the identity. Now suppose it is not injective. We get there exists $y\neq1$ such that $f(y)=\frac{1}{y}$. We now use the other condition to show that this implies $f$ is the reciprocal function. Let $x\neq1$ be arbitrary. WLOG, we can take $y>1$, $x<1$ and $x>\frac{1}{y}>0$ (replacing $x$ and $y$ by their reciprocals and powers if necessary). We get $1\leq xyf(xy)=xf(x)\leq1$. Hence $f(x)=\frac{1}{x}$ for all $x\in\mathbb{R}_{>0}$ and $f$ is the reciprocal function. Remarks: It seems if you ignored the inequality condition, the problem would be purely algebraic (together with knowing squaring is one to one) and the functions would be described by the following decomposition: Let $\mathbb{R}_{>0}^{f=1/id}$ denote the subset on which $f$ is the reciprocal. Then $$\mathbb{R}_{>0}=\mathbb{R}_{>0}^{f=id}\times\mathbb{R}_{>0}^{f=1/id}.$$(This is actually a decomposition of abelian groups where the factors are the image and kernel of $g$ which is a homomorphism.) So if you wanted to solve for $f$ without the inequality condition it would be equivalent to all ways of finding two multiplicative subsets whose intersection is $\{1\}$ and together generate $\mathbb{R}_{>0}$ under multiplication.

Unfortunately, this was a bit bashy. The only solutions are $a\diamondsuit b = ab$ and $a\diamondsuit b = a/b$. It is easy to check they work. We now show these are the only solutions. Consider some solution $\diamondsuit$. Define $f:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ be defined by $f(x) = 1\diamondsuit x$ (this is a definition that will help us later on). Claim: We have $1\diamondsuit 1 = 1$. Solution: Remark that \[(1\diamondsuit 1) \cdot (1\diamondsuit 1) = 1\diamondsuit (1\diamondsuit (1\diamondsuit 1)) = 1\diamondsuit ((1\diamondsuit 1)\cdot 1) = 1\diamondsuit (1\diamondsuit 1) = (1\diamondsuit 1) \cdot 1 = 1\diamondsuit 1.\]As $1\diamondsuit 1 > 0$, this implies the desired. $\fbox{}$ Remark that for any $x$, \[f(f(x))=1\diamondsuit (1\diamondsuit x) = (1\diamondsuit 1)\cdot x = 1\cdot x = x.\]Plug in $b=1,c=1\diamondsuit x$ to get \[a\diamondsuit x = a\diamondsuit (1 \diamondsuit (1\diamondsuit x)) = a\cdot (1\diamondsuit x) = af(x).\]This prompts us to rewrite the original given information as \[af(bf(c)) = af(b) \cdot c,\]or \[g(bf(c))=cf(b).\]We additionally have $f(f(x))=x$ for all $x\in \mathbb{R}_{\ge 0}$. Note that this implies $f$ is bijective, so the original statement is equivalent to \[f(bc) = f(b)f(c)\]for all $b,c$. Hence, $g(x)=\log f(e^x)$ satisfies the Cauchy equation. Note that by the second given piece of information, we have $f(a) \ge 1/a$ for all $a\ge 1$. In particular, this means that $g$ has the lower bound of $\log 1/2$ on the interval $[\log 1,\log 2]$. Hence, $g(x)$ is linear and $f(x)$ is an exponential function $x^c$ for some real $c$. By the equation $f(f(x))=x$, we have $x^{c^2} = x$, so $c\in \{-1,1\}$. Hence, by $a\diamondsuit b = af(b)$, we get the claimed solutions of $ab,a/b$.

Since I don't like diamonds, denote $f(a,b)=a\,\diamondsuit\, b$, so that our conditions become \[ f(a,f(b,c))=cf(a,b) \qquad\text{ and } \qquad f(a,a)\geq 1, \text{ when } a\geq 1\] The only functions which work are $f(a,b)=ab$ and $f(a,b)=a/b$, which clearly work. The main idea is to show $g(x)=f(1,x)$ is a multiplicative, determining $f$ completely. As usual, let $P(a,b,c)$ denote the assertion $f(a,f(b,c))=cf(a,b)$. Claim: $f$ is 'injective', that is, $f(a,x)=f(a,y)$ and $f(x,a)=f(y,a)$ both imply $x=y$. The former is immediate by comparing $P(a,b,x)$ and $P(a,b,y)$. For the latter, observe that $P(a,b,1)\implies f(b,1)=b$ by this `injectivity'. Then comparing $P(x,1,c)$ and $P(y,1,c)$ gives our other form of `injectivity'. Claim: $af(1,b)=f(a,b)$ $P(a,1,c)$ implies \[ f(a,f(1,c))=ac\]Then by putting $c=\frac{1}{a}$ in the above, we obtain $f(a,f(,\frac{1}{a}))=1$. However, $P(a,b,\frac{1}{f(a,b)})\implies f(a,f(b,\frac{1}{f(a,b)}))=1$ from which injectivity gives $f(b,\frac{1}{f(a,b)})=f(1,\frac{1}{a})$. Now since $f(b,1)=b$, $f(1,1)=1$. Hence letting $a=1$ in this gives $f(b,\frac{1}{f(1,b)})=1$. Finally, $P(a,b,\frac{1}{f(1,b)})$ gives\begin{align*} f(a,f(b,\frac{1}{f(1,b)}))&=\frac{f(a,b)}{f(1,b)} \\ \iff f(a,1)&=\frac{f(a,b)}{f(1,b)} \\ \iff af(1,b) &= f(a,b) \end{align*}as required. Claim: $g(x)=f(1,x)$ is multiplicative $P(1,b,c)$ gives \[ f(1,f(b,c)) = cf(1,b) = f(c,b) \]by the previous claim. In the above, put $b=\frac{1}{f(1,c)}$ so that \[f\left(1,f\left(\frac{1}{f(1,c)},c\right)\right)=f(c,\frac{1}{f(1,c)})=1=f(1,1)\]giving $f\left(\frac{1}{f(1,c)},c\right)=1$ by injectivity. Recall that $f(b,\frac{1}{f(1,b)})=1$ - setting $b=\frac{1}{f(1,c)}$ gives, after some computation, $f(1,\frac{1}{f(1,c)})=\frac{1}{c}$. Finally, \begin{align*} P(1,b,\frac{1}{f(1,c)})\implies f(1,f(b,\frac{1}{f(1,c)}))&= \frac{f(1,b)}{f(1,c)} \\ \iff f(1,bf(1,\frac{1}{f(1,c)}))&= \frac{f(1,b)}{f(1,c)} \\ \iff f(1,c)f(1,\frac{b}{c})&=f(1,b) \end{align*}hence $g$ is multiplicative as claimed. Now let $h(x)=\log g(e^x)$, so that $g(xy)=g(x)g(y)\implies h(x+y)=h(x)+h(y)$. The second condition gives $g(x)\geq \frac{1}{x}$ for all $x\geq 1$. Then for any $x\geq 0$, $e^x\geq 1$ so $h(x)=\log g(e^x)\geq \log e^{-x} = -x$. Since any non-linear additive function must be dense in the plane, it follows that $h$ is linear, from which we deduce $g(x)=x^k$ for some constant $k$. Finally, by expanding out $P(1,1,c)$, we find that $c^{k^2}=c$, so $k=1$ or $k=-1$ which correspond to the claimed solutions.

Really nice problem! Replace $a\diamondsuit b$ with $f(a,b),$ and let $P(a,b,c)$ denote the first assertion. $\textbf{Claim: }$ For fixed $a,$ the function $f(a,b)$ is injective. $\emph{Proof: }$ Suppose $f(a,b_1)=f(a,b_2).$ Then, $P(c,a,b_1)$ and $P(c,a,b_2)$ give $$b_{1}f(c,a)=f(c,f(a,b_1))=f(c,f(a,b_2))=b_{2}f(c,a).$$Since $f(c,a)\ne 0,$ we must have $b_1=b_2,$ as needed. $\blacksquare$ Now define a function $g:\mathbb{R^+}\to\mathbb{R^+}$ such that $g(a)=f(1,a)$ for all $a.$ $\textbf{Claim: }$ $f(a,1)=a$ for all $a.$ $\emph{Proof: }$ From $P(b,a,1),$ we have $$f(b,f(a,1))=f(b,a),$$so we are done by injectivity. $\blacksquare$ $\textbf{Claim: }$ $f(a,b)=ag(b)$ for all $a,b.$ $\emph{Proof: }$ $P(a,b,\frac{a}{f(a,b)})$ yields $$f\left(a,f\left(b,\frac{a}{f(a,b)}\right)\right)=a\implies f\left(b,\frac{a}{f(a,b)}\right)=1.$$On the other hand, $P(1,b,\frac{1}{f(1,b)})$ gives $$f\left(1,f\left(b,\frac{1}{f(1,b)}\right)\right)=1\implies f\left(b,\frac{1}{f(1,b)}\right)=1.$$Therefore, $\frac{a}{f(a,b)}=\frac{1}{f(1,b)}$ for all $a,b.$ This implies that $f(a,b)=af(1,b)=ag(b),$ as desired. $\blacksquare$ Using the above claim, $P(1,1,c)$ yields $g(g(c))=c$ and $P(1,b,g(c))$ yields $g(bc)=g(b)g(c).$ Now we have two cases: $\textbf{Case 1: }$ There exists $a>1$ such that $g(a)<1.$ By the given inequality, we know $g(a)\ge\frac{1}{a}.$ Furthermore, we have $$a=g(g(a))=\frac{1}{g(\frac{1}{g(a)})}\le\frac{1}{g(a)}\implies g(a)\le\frac{1}{a}.$$Therefore, we have $g(a)=\frac{1}{a}.$ Now consider some $x\in\mathbb{R}^+.$ Let $k$ be the integer such that $a^k\le x<a^{k+1}.$ Since $g$ is multiplicative, we have $g(a^k)=\frac{1}{a^k}$ and $g(a^{k+1})=\frac{1}{a^{k+1}}.$ Thus, $$g(x)=\frac{1}{a^k}g\left(\frac{x}{a^k}\right)\ge\frac{1}{x},$$$$g(x)=\frac{\frac{1}{a^{k+1}}}{g\left(\frac{a^{k+1}}{x}\right)}\le\frac{1}{x}.$$Hence, $g(x)=\frac{1}{x}$ for all $x\in\mathbb{R}^+$ in this case. $\square$ $\textbf{Case 2: }$ If $a>1,$ then $g(a)>1.$ Since $g$ is multiplicative, the condition implies that $g$ is strictly increasing. Therefore, since $g$ is an involution, it must be the identity. $\square$ The solutions above yield $\boxed{a\diamondsuit b=ab}$ and $\boxed{a\diamondsuit b=a/b},$ which can be easily checked to work.

Define $f(a,b)=a \ \diamondsuit \ b$. The condition is \[ P(a,b,c):\quad f(a,f(b,c))=f(a,b)\cdot c. \]The key is to analyze $g(x) := f(1,x)$. Claim: $g(1)=1$. Proof: We will show $f(1,1)=1$. We have \begin{align*} f(a,b) &= f(a,f(b,1)) \qquad \qquad \text{by } P(a,b,1) \\ &= f(a,f(b,f(1,1))) \quad \ \text{by } P(b,1,1)\\ &= f(a,b)\cdot f(1,1) \qquad \ \ \text{by } P(a,b,f(1,1)). \end{align*}Since $f$ is positive, this implies $f(1,1)=1$. $\blacksquare$ Claim: $g$ is an involution, and hence a bijection. Proof: $P(1,1,x)$ implies $f(1,f(1,x))=f(1,1)x$, i.e. $g(g(x))=x$. Therefore, $g$ is an involution, which means it is also a bijection. $\blacksquare$ Claim: $g$ is multiplicative. Proof: Firstly, by $P(1,b,1)$, we have $f(1,f(b,1))=f(1,b)$, so $g(f(b,1))=g(b)$. Since $g$ is bijective, this implies $f(b,1)=b$ for all $b$. Note that $f(b,g(c))=f(b,f(1,c))=f(b,1)c=bc$ where we used $P(b,1,c)$ in the middle. So now, $P(1,b,g(c))$ implies $g(f(b,g(c)))=g(b)g(c)$, i.e. $g(bc)=g(b)g(c)$. $\blacksquare$ Now, let $h(x)=\log g(e^x)$. Since $g$ is multiplicative, $h$ is additive. We want to involve the second condition, and bring in $f(a,a)$ somehow. We have \[ P(a,1,g(a)):\quad f(a,g(g(a))=ag(a) \implies f(a,a) = ag(a) \implies g(a)=\tfrac{1}{a}f(a,a) \ge \tfrac{1}{a} \]for $a\ge 1$. Therefore, $h(x) = \log g(e^x) \ge \log e^{-x} = -x$. In particular, $h$ is bounded below on some nontrivial interval. Combined with $h$ being additive, we have $h(x)=kx$ for some constant $k \in \mathbb{R}$. Now, $g(e^x)=e^{kx}$, so $g(x)=x^k$. From $P(a,1,c)$, we have $f(a,g(c))=ac$, so now $f(a,c^k)=ac$, so $f(a,d)=ad^{1/k}$, so $f(a,b)=ab^\ell$ for some constant $\ell$. Plugging this into $P(a,b,c)$ gives $a(bc^\ell)^\ell=ab^kc\implies \ell^2=1 \implies \ell=\pm 1$. Therefore, $f(a,b)=ab$ or $f(a,b)=a/b$. Both are easy to confirm that they work.

Sad algebra. Let $P(a,b,c)$ denote the first condition. The answer is multiplication and division, which clearly work. $P(a,b,1)$ gives $a \diamondsuit (b \diamondsuit 1)=a \diamondsuit b$. Left-multiplying by $a$ and applying the given condition demonstrates $b \diamondsuit 1=b$. Taking $P(1,b,1 \diamondsuit c)$ gives$$1 \diamondsuit (b \diamondsuit (1 \diamondsuit c))= (1 \diamondsuit b)(1 \diamondsuit c)$$but since $(b \diamondsuit (1 \diamondsuit c))=bc$ we actually have $(1 \diamondsuit bc)=(1 \diamondsuit b)(1 \diamondsuit c)$, i.e. the function $f(x):=1 \diamondsuit x$ is multiplicative. Also note $P(1,1,c)$ tells us $f$ is involutive. Now do the $g(x):= \ln (f(e^x))$ transformation, so $g: \mathbb{R} \to \mathbb{R}$ is Cauchy. We show $g(x) \ge -x \forall x \ge 0$, hence $g$ is linear. Indeed, it's equivalent to $f(x) \ge \frac{1}{x} \forall x \ge 1$, which follows from$$1 \le a \diamondsuit a = a \diamondsuit (1 \diamondsuit (1 \diamondsuit a)) = a(1 \diamondsuit a).$$Hence $g(x)=dx$ for some constant $d$ and $f(x)=x^d$. But since $f$ is involutive we have $d= \pm 1$. $P(1,b,c)$ now gives the desired conclusion.

Reminds me of 2016 amc 10a #23. Let $P(a,b,c)$ denote the given assertion. Claim: $\diamondsuit$ is "injective" (if $a\diamondsuit b =a\diamondsuit c$, then $b=c$). Proof: $P(a,a,b): a\diamondsuit(a\diamondsuit b)=(a\diamondsuit a)\cdot b$. $P(a,a,c): a\diamondsuit(a\diamondsuit b)=(a\diamondsuit a)\cdot c$. Since $a\diamondsuit a\ne0$, $b=c$. $P(a,a,1): a\diamondsuit(a\diamondsuit 1)=a\diamondsuit a$, so $a\diamondsuit 1=a$. $P(a,1,b): a\diamondsuit (1\diamondsuit b)=a\cdot b$. $P(1,1,b): 1\diamondsuit (1\diamondsuit b)=b$. $P(a,1, 1\diamondsuit b): a\diamondsuit b=a\cdot (1\diamondsuit b)$. $P(1,a,1\diamondsuit b): 1\diamondsuit (a\diamondsuit (1\diamondsuit b))=1\diamondsuit (a\cdot (1\diamondsuit (1\diamondsuit b)))=1\diamondsuit (a\cdot b)=(1\diamondsuit a)\cdot (1\diamondsuit b)$. So the function $f(x)=1\diamondsuit x$ is multiplicative. Now we take $g(x)=\ln(f(e^x))$ so $g$ is additive. Claim: $g$ is bounded over the interval $(0,1)$. Proof: If $a\ge1$, then \[f(a)=\frac{1}{a}\cdot (a\diamondsuit a)\ge \frac{1}{a}\] Set $e^x=a$, so $1<a<e$. We have \[g(x)=\ln(f(a))\ge \ln(\frac{1}{a})>-1,\]which proves our claim. This implies $g(x)=kx$, so $f(x)=x^k$. Claim: $k=\pm1$. Proof: $P(1,a,b): f(a\cdot f(b))=f(a)\cdot b=a^k\cdot b^{(k^2)}=a^k\cdot b$. Since $a\ne0$, we have $b^{k^2}=b\implies k^2=1\implies k=\pm1$. If $k=1$, then we have $\boxed{a\diamondsuit b=a\cdot b}$. If $k=-1$, then we have $\boxed{a\diamondsuit b=\frac{a}{b}}$. In other words, the only two such operations are multiplication and division, which both work.

We should have more binary operation FE. The answer is $a \,\diamondsuit\, b = ab$ and $a \,\diamondsuit\, b = \tfrac{a}{b}$ which clearly work. Now we show that these are the only functions. Denote the first assertion by $P(a,b,c)$. Suppose that $a \,\diamondsuit\, b = a \,\diamondsuit\, c$, so by comparing $P(a,a,b),P(a,a,c)$ we have $$(a \,\diamondsuit\, a)b=a \,\diamondsuit\, (a \,\diamondsuit\, b)=a \,\diamondsuit\, (a \,\diamondsuit\, c)=(a \,\diamondsuit\, a)c,$$thus $b=c$. Thus $\,\diamondsuit\,$ is "injective". Now from $P(a,a,1)$ we have $$a \,\diamondsuit\, (a \,\diamondsuit\, 1)=a \,\diamondsuit\, a \implies a \,\diamondsuit\, 1 = a$$by "injectivity". Next, $P(a,1,c)$ gives $$a \,\diamondsuit\, (1 \,\diamondsuit\, c)=ac,$$in particular giving $1 \,\diamondsuit\, (1 \,\diamondsuit\, a))=a$, so from $P(a,1,1 \,\diamondsuit\, c)$ we obtain $$a \,\diamondsuit\, c=a \,\diamondsuit\, (1 \,\diamondsuit\, (1 \,\diamondsuit\, c))=(a \,\diamondsuit\, 1)(1 \,\diamondsuit\, c)=a(1 \,\diamondsuit\, c).$$Then $P(1,b,c)$ gives $$1 \,\diamondsuit\, (b(1\,\diamondsuit\, c))=1\,\diamondsuit\, (b \,\diamondsuit\, c)=c(1 \,\diamondsuit\, b).$$Letting $f(x)=1\,\diamondsuit\, x$, this means that $f(bf(c))=cf(b)$. Further $f$ is bijective and thus surjective as $f(f(a))=a$, so letting $a=f(c)$ we get $f(ab)=f(a)f(b)$ for all $a,b \in \mathbb{R^+}$. Now let $g(x)=\ln(f(e^x))$, so $g$ is $\mathbb{R} \to \mathbb{R}$ and additive. Further, for $a \in [1,e]$ we have $$a(1\,\diamondsuit\, a)=a \,\diamondsuit\, a\geq 1 \implies f(a)\geq \frac{1}{e} \implies g(\log(a))\geq -1,$$so for $x \in [0,1]$, $g(x)$ is bounded below, thus by Cauchy we have $g(x)=kx$ for $k \in \mathbb{R}$. Substituting, this yields $f(x)=x^k$, so $$a \,\diamondsuit\, b=a(1\,\diamondsuit\, b)=ab^k.$$To finish, $P(a,b,c)$ yields $$ab^kc^{2k}=ab^kc \implies k^2=1 \implies k=\pm 1,$$hence $a \,\diamondsuit\, b=ab$ or $a \,\diamondsuit\, b=\tfrac{a}{b}$. $\blacksquare$

The solutions are $a \,\diamondsuit\, b = ab$ and $a \,\diamondsuit\, b = \frac{a}{b}$. Claim 1: For any fixed $b$, the function $h_b(x) := b \,\diamondsuit \, x$ is injective. Proof: Suppose there are $x_1, x_2$ such that $h_b(x_1) = h_b(x_2)$. Then, $$x_1 = \frac{a \,\diamondsuit\, (b \,\diamondsuit \, x_1)}{a \,\diamondsuit b} = \frac{a \,\diamondsuit\, (b \,\diamondsuit \, x_2)}{a \,\diamondsuit b} = x_2$$$\Box$ We then have $$a \,\diamondsuit\, (b \,\diamondsuit\, 1) = a \,\diamondsuit\, b \qquad \Rightarrow \qquad b \,\diamondsuit \, 1 = b$$Define $f(x) := 1 \,\diamondsuit \, x$. Claim 2: We have $x \,\diamondsuit\, y = x \cdot f(y)$. The function $f$ also satisfies $f(f(x)) = x$, and $f(xy) = f(x) \cdot f(y)$. Proof: Note that $f(1) = 1$ and $$a \,\diamondsuit\, f(c) = (a \,\diamondsuit\, 1) \cdot c = ac$$Setting $a = 1$ gives us $f(f(c)) = c$. We can then characterize the $\diamondsuit$ operation as $$a \,\diamondsuit\, (1 \,\diamondsuit\, f(b)) = (a \,\diamondsuit\, 1) \cdot f(b) \qquad \Rightarrow \qquad a\,\diamondsuit\, b = a\cdot f(b)$$Finally, $$1 \,\diamondsuit\, (f(b) \,\diamondsuit\, c) = (1 \,\diamondsuit\, f(b)) \cdot c \qquad \Rightarrow \qquad f(f(b) \cdot f(c)) = bc$$Taking $f$ of both sides yields $f(bc) = f(b) \cdot f(c)$, as desired. $\Box$ Defining $g(x) := \ln f(e^x)$, our findings translate to the following $g(x + y) = g(x) + g(y)$ $g(g(x)) = x$ $g(x) + x \ge 0$ for $x \ge 0$. The first is the famous Cauchy functional equation. If we define $g_1(x) := g(x) + x$, notice that $g_1(x + y) = g_1(x) + g_1(y)$ as well. Given that $g_1$ is linear and bounded below at $0$ for $x \ge 0$, it is well known that $g_1(x) = (k - 1)x$ for some constant $k$, and thus $g(x) = kx$. Of course, the only $k$ that satisfy condition 2 is $k = \pm 1$, so $g(x) = x$ or $g(x) = -x$. These solutions translate into $f(x) = x$ and $f(x) = \frac{1}{x}$, which gives us the final solution of $ab$ and $\frac{a}{b}$, both of which can be checked to work.

Replace $a\,\diamondsuit\,b$ with $f(a,b)$, because when I was solving the problem it was too hard to write a $\diamondsuit$ symbol. We first prove that $f$ is right injective, or that $f(b,c_1)=f(b,c_2)$ means that $c_1=c_2$. Assume this is not the case, and such $b,c_1,c_2$ exist. Then $f(a,b)\cdot c_1=f(a,f(b,c_1))=f(a,f(b,c_2))=f(a,b)\cdot c_2$, which is impossible. $f(a,f(a,1))=f(a,a)$, so $f(a,1)=a$. $f\left(a,f\left(b,\frac{a}{f(a,b)}\right)\right)=a$, so $f\left(b,\frac{a}{f(a,b)}\right)=1$. Thus $\frac{a}{f(a,b)}$ is fixed as $a$ varies, so $\frac{a}{f(a,b)}=\frac{b}{f(b,b)}$ and $f(a,b)=\frac{a}{b}\cdot f(b,b)$. Let $g(b)=\frac{f(b,b)}{b}$. Rewriting the given equation in terms of $g$ eventually gives $g(bg(c))=g(b)\cdot c$ for any $b,c$. In addition, $f(1,1)=1$, so $g(1)=1$ and $g(g(c))=c$. Thus we can rewrite the given as $g(bc)=g(b)g(c)$. $g(b)\ge \frac{1}{b}$ for $b\ge 1$, so $g(b)=b^{\text{const}}$. Since $g(g(b))=b$, $g(b)=\frac{1}{b}$ or $g(b)=b$, and $f(a,b)$ either equals $ab$ for all $a,b$ or $\frac{a}{b}$ for all $a,b$, as desired.

the hardest part of this problem was drawing the diamond. Let $P(a,b,c)$ denote the given assertion, and because i'm lazy replace $a \, \diamondsuit \, b$ with $f(a,b)$. $P(1, 1, 1) \implies f(1, 1) = f(1, f(1, 1))$. $P(1, 1, f(1, 1)) \implies f(1, 1) \cdot f(1, 1) = f(1, f(1, f(1, 1))) = f(1, f(1, 1)) = f(1, 1) \implies f(1, 1) = 1$. $P(1, 1, c) \implies f(1, f(1, c)) = c$; defining $g(x)= f(1, x)$ means that this can be rewritten as $g(g(x)) = x$. $P(a, 1, g(c)) \implies f(a, c) = ag(c)$. Using this equation, rewrite the given condition as \[ag(bg(c)) = ag(b) \cdot c \implies g(bg(c)) = cg(b).\]Since $g(g(x)) = x$, $g$ is bijective, so therefore we can rewrite this equation as $g(xy) = g(x)g(y)$. Letting $h(x) = \log g(e^x)$ tells us that $h$ satisfies Cauchy. The second piece of information tells us that in fact, $g(a) \ge 1/a$ for $a \ge 1 \implies h(x)$ is bounded below by $\log 1/e = -1$ over $[\log 1,\log e] = [0, 1] \implies h$ linear $\implies g = x^k$ for some $k$. However, note that \[g(g(x)) = x \implies k = \pm 1 \implies f(a,b) = ab \,\, \text{or}\,\, a/b,\]which both work.

Note that $a \, \diamondsuit \, b = ab$ and $a \, \diamondsuit \, b = a/b$ are both satisfactory definitions for $\diamondsuit$. We will now show that they are the only ones. Denote the first assertion by $P(a, b, c)$. We start off by analyzing the function $f \colon x \mapsto 1 \, \diamondsuit \, x$. $P\left(1, b, \frac{x}{1 \, \diamondsuit \, b}\right)$ gives $1 \, \diamondsuit \, \left(b \, \diamondsuit \, \frac{x}{1 \, \diamondsuit \, b}\right) = x$. Hence, $f$ is surjective. Suppose that $1 \, \diamondsuit \, a = 1 \, \diamondsuit \, b$. Then, \[(1 \, \diamondsuit \, 1) \cdot a = 1 \, \diamondsuit \, (1 \, \diamondsuit \, a) = 1 \, \diamondsuit \, (1 \, \diamondsuit \, b) = (1 \, \diamondsuit \, 1) \cdot b,\]so $a = b$. $f$ is injective. We now systematically plug in triples $(a, b, c)$ with as many $1$s as we like, progressively gaining a foothold on the structure of $f$. Let $x$ and $y$ be arbitrary positive real numbers. $P(1, 1, 1) \implies 1 \, \diamondsuit \, (1 \, \diamondsuit \, 1) = 1 \, \diamondsuit \, 1 \implies 1 \, \diamondsuit \, 1 = 1 \implies f(1) = 1$. $P(1, 1, x) \implies 1 \, \diamondsuit \, (1 \, \diamondsuit \, x) = x \implies f(f(x)) = x$. $P(1, x, 1) \implies 1 \, \diamondsuit \, (x \, \diamondsuit \, 1) = 1 \, \diamondsuit \, x \implies x \, \diamondsuit \, 1 = x$. $P(1, x, y) \implies 1 \, \diamondsuit \, (x \, \diamondsuit \, y) = (1 \, \diamondsuit \, x) \cdot y \implies f(x \, \diamondsuit \, y) = f(x) \cdot y$. $P(x, 1, y) \implies x \, \diamondsuit \, (1 \, \diamondsuit \, y) = xy \implies x \, \diamondsuit \, f(y) = xy \implies x \, \diamondsuit \, y = xf(y)$. Combining the last two equations, we obtain $f(xf(y)) = f(x)y$, alternatively $f(xy) = f(x)f(y)$. We define the helper function $g(z) = \log f(e^z)$ for all $z \in \mathbb{R}$. Then, $g(x) + g(y) = g(x + y)$, i.e. $g$ is a Cauchy function. The second given assertion is equivalent to $af(a) \geq 1$ for all $a \geq 1$, and this statement translates to $z + g(z) \geq 0$ for all $z \geq 0$. Hence, $g$ is not dense in the plane; it must be linear. Checking all such $g$ verifies that the only possibilities are $f \equiv 1/x$ and $f \equiv x$. We obtain our two solutions from there.

The answer is multiplication or division. Claim: everything's bijective Proof: If we fix the left operand, we can obtain every value by varying the right operand: this can be seen by fixing $a$ and $b$ and varying $c$ in the first equation. Furthermore, if we fix the left operand, every value of the right operand gives a distinct value for the expression: this can be also be seen by varying $c$ in the first equation, since $(b \diamondsuit c)$ must be distinct for each distinct $c$. In other words, $f(x) = C \diamondsuit x$ is a bijection for any positive constant $C$. By fixing $a$ and $c$ and varying $b$, we see similarly that $f(x) = x \diamondsuit C$ is a bijection for any positive constant $C$. Because of this, by varying $b$ in the first equation, it follows that for any fixed constant $C$, the function $f(x) = x \diamondsuit C$ is also injective. In fact, by varying $c$, it follows that $f(x) = x \diamondsuit C$ is bijective. Plugging in $c=1$ gives us that $b \diamondsuit 1 = b$, and plugging in $b=1$ gives us that $a \diamondsuit (1 \diamondsuit c) = ac$. Hence, the function $f(x) = x \diamondsuit C$ is linear for any positive constant $C$. Plugging in $a=1$ and letting $f(x) = 1 \diamond x$, we can manipulate the given equation: \begin{align*} 1 \diamond (b \diamond c) &= f(b) \cdot c \\ 1 \diamond (b ( 1 \diamond c)) &= f(b) \cdot c \\ f(bf(c)) &= f(b) \cdot c. \end{align*}Plugging in $b=1$ gives $f(f(c)) = c$, and plugging $c = f(c)$ gives \[ f(bc) = f(b)f(c). \]Letting $g(x) = \ln \left( f \left( e^x \right) \right)$, we have $g(a+b) = g(a) + g(b)$. The second condition in the problem bounds $g$, hence $g(x) = cx$, so $f(x) = x^c$ for some $c$. (if that wasn't a fakesolve my mind is blown....) Hence $a \diamond b$ is $ab^C$ for some constant $C$. Plugging back into the first equation forces $C = \pm 1$.

Let $P(a, b, c)$ denote the given assertion and let $f(a, b) = (a \diamondsuit b)$. Consider the following claim: Claim: $f(a, 1) = a$ for all $a \in \mathbb{R}_{>0}$. Proof. $P(1, b, 1)$ gives $f(1, f(b, 1)) = f(1, b)$ and $P(a, 1, c)$ gives $f(a, 1) \cdot c = f(a, f(1, c)) = f(a, f(1, f(c, 1))) = f(a, 1) \cdot f(c, 1)$, thus $f(c, 1) = c$ for all $c \in \mathbb{R}_{>0}$. $\blacksquare$ Claim: $f(a, b) = a \cdot f(1, b)$ for all $a, b \in \mathbb{R}_{>0}$. Proof. Note that $f(a, b) = f(a, f(1, 1)b) = f(a, f(1, f(1, b))) = f(a, 1) \cdot f(1, b) = f(1, b) \cdot a$, as needed. $\blacksquare$ Now let $g(x) = f(1, x)$. Our given functional equation rewrites to $g(bg(c)) = g(b)c$. Taking $b = 1$ gives $g(g(c)) = g(1)c = c$, thus $g$ is an involution. Therefore $g$ is injective, so plugging $g(c)$ as $c$ gives $g(bc) = g(b)g(c)$, so $g$ is multiplicative. Let $h(x) = \ln g(e^x)$, then $h(x + y) = h(x) + h(y)$ and second condition becomes $h(x) \ge -x$. Thus $h$ is bounded below on interval $(0, 1)$ thus $h(x) = kx$. Since $g(g(x)) = x$ implies $h(x) = x$ or $h(x) = -x$. Thus $g(x) = x$ or $g(x) = \frac{1}{x}$ and hence $f(a, b) = ab$ or $f(a, b) = \frac{a}{b}$. Both answers satisfy the condition, so we're done. $\blacksquare$

Replace the diamond with $f$ so that $$f(a,f(b,c))=cf(a,b).$$We claim the answer is $f(a,b)=ab$ or $f(a,b)=\frac{a}{b}.$ Claim: If $f(b,c)=b$, then we must have $c=1$. Plug in any $c$ such that $f(b,c)=b$. Then, the equation becomes $$f(a,b)=cf(a,b),$$so we must have $c=1$. We can actually do a lot with this. For example, plugging $c=\frac{a}{f(a,b)}$, we have $$f(a,f(b,\frac{a}{f(a,b)}))=a,$$which means that $$f(b,\frac{a}{f(a,b)})=1 (*).$$ Now, plugging $a=b=1$ here, we have $$f(1,\frac{1}{f(1,1)})=1,$$which means that $$\frac{1}{f(1,1)}=1$$$$f(1,1)=1.$$ Claim: $g_a(x)=f(a,x)$ is injective for all $a$. Note that $$c=\frac{f(a,g_b(c))}{f(a,b)}.$$Since $c$ is uniquely determined by $g_b(c)$, this shows the claim. We further claim that $g_a$ is also surjective. Consider the original equation $$g_a(g_b(c))=cf(a,b).$$As we fix $a,b$ and vary $c$, the LHS can take on any positive real. Thus, $f(a,x)$ is bijective for a fixed $a$. By $(*)$ as well as bijectivity, for any $b$, the corresponding second input that results in a value of 1 is always $$\frac{a}{f(a,b)}.$$In particular, this is constant for all $a$. Thus, $\frac{a}{f(a,b)}$ is constant, which means that $$f(a,b)=ah(b)$$for some function $h$. Rewrite the original functional equation in terms of $h$ so that $$h(bh(c))=ch(b)$$and $$h(x)\geq \frac{1}{x}$$for $x\geq 1.$ We know that $h(1)=1$. Plugging $b=1$ we see that $h$ is an involution. Thus, $b\rightarrow f(b)$ gives $$h(h(b)h(c))=bc$$$$h(b)h(c)=h(bc).$$The solution to this over positive reals is $h(x)=x^\alpha$ (to do this take ln and substitute $v(x)=\ln h(e^x).$, the constraint $h(x)\geq \frac{1}{x}$ gives the necessary bound for Cauchy to work). However, plugging that back into $h(bh(c))=ch(b)$, we see that only $\alpha=\pm 1$ work, done.

We claim that $\boxed{a\,\diamondsuit\,b\equiv a\cdot b}$ or $\boxed{a\,\diamondsuit\,b\equiv \frac{a}{b}}$. Clearly these satisfy the conditions. We have \[ (1\,\diamondsuit\,1)(1\,\diamondsuit\,1)=1\,\diamondsuit\,(1\,\diamondsuit\,(1\,\diamondsuit\,1))=1\,\diamondsuit\,(1\,\diamondsuit\,1)=1\,\diamondsuit\,1 \]so $1\,\diamondsuit\,1=1$. Claim 1. If $a\,\diamondsuit\,x=a\,\diamondsuit\,y$ then $x=y$. Proof. We have \[ (1\,\diamondsuit\,a)x=1\,\diamondsuit\,(a\,\diamondsuit\,x)=1\,\diamondsuit\,(a\,\diamondsuit\,y)=(1\,\diamondsuit\,a)y \]so $x=y$. $\square$ Claim 2. We have $a\,\diamondsuit\,1=a$ for all $a\in\mathbb{R}^+$. Proof. We have \[ 1\,\diamondsuit\,(a\,\diamondsuit\,1)=1\,\diamondsuit\,a \]so $a\,\diamondsuit\,1=a$ by Claim 1. $\square$ Claim 3. For $a\in\mathbb{R}^+$, there exists a unique $\widehat{a}$ such that $a\,\diamondsuit\,\widehat{a}=1$. Moreover, $\widehat{a}=1\,\diamondsuit\,\frac{1}{a}$. Proof. Note that \[ a\,\diamondsuit\,\left(1\,\diamondsuit\,\frac{1}{a}\right)=(a\,\diamondsuit\,1)\cdot\frac{1}{a}=1 \]by Claim 2. $\widehat{a}$ is unique by Claim 1. $\square$ Let $f(x):=1\,\diamondsuit\,x$. Then \begin{align*} f(a)f(b)&=(1\,\diamondsuit\,a)(1\,\diamondsuit\,b)\\ &=1\,\diamondsuit\,(a\,\diamondsuit\,(1\,\diamondsuit\,b))\\ &=1\,\diamondsuit\,((a\,\diamondsuit\,1)\cdot b)\\ &=1\,\diamondsuit\,ab\\ &=f(ab) \end{align*}for all $a,b\in\mathbb{R}^+$. We also have \[ f(f(x))=1\,\diamondsuit\,(1\,\diamondsuit\,x)=(1\,\diamondsuit\,1)\cdot x=x. \]Let $g(x):=\log f(e^x)$, which is defined because $f(x)$ is always positive. Then $g(x+y)=g(x)+g(y)$ for all $x,y\in\mathbb{R}$. Note that \[ a\,\diamondsuit\,\left(a\,\diamondsuit\,\frac{1}{a\,\diamondsuit\,a}\right)=(a\,\diamondsuit\,a)\cdot\frac{1}{a\,\diamondsuit\,a}=1 \]so $a\,\diamondsuit\,\frac{1}{a\,\diamondsuit\,a}=\widehat{a}=1\,\diamondsuit\,\frac{1}{a}$ by Claim 3. Then \[ \frac{1\,\diamondsuit\,a}{a\,\diamondsuit\,a}=1\,\diamondsuit\,\left(a\,\diamondsuit\,\frac{1}{a\,\diamondsuit\,a}\right)=1\,\diamondsuit\,\left(1\,\diamondsuit\,\frac{1}{a}\right)=\frac{1}{a} \]so $f(a)=1\,\diamondsuit\,a=\frac{a\,\diamondsuit\,a}{a}\geq\frac{1}{a}$ for $a\geq 1$. Thus we have \[ g(x)=\log f(e^x)\geq\log\frac{1}{e^x}=-x \]for all $x\geq 0$. It follows that $g$ is bounded in the interval $[0,1434]$ so $g(x)=cx$ for some $c\in\mathbb{R}$. Thus $f(x)=x^c$. Since $f(f(x))=x$, we must have $c=\pm 1$ so $1\,\diamondsuit\,x\equiv x$ or $1\,\diamondsuit\,x\equiv\frac{1}{x}$. If $1\,\diamondsuit\,x\equiv x$ then \[ a\,\diamondsuit\,b=1\,\diamondsuit\,(a\,\diamondsuit\,b)=(1\,\diamondsuit\,a)\cdot b=a\cdot b. \]If $1\,\diamondsuit\,x\equiv\frac{1}{x}$ then \[ a\,\diamondsuit\,b=\frac{1}{1\,\diamondsuit\,(a\,\diamondsuit\,b)}=\frac{1}{(1\,\diamondsuit\,a)\cdot b}=\frac{a}{b}. \]We are done. $\square$

Really nice and original problem. I found it pretty instructive. The answers are $a\,\diamondsuit\,b = ab$ for all $(a,b)\in \mathbb R_{>0}\times \mathbb R_{>0}$ and $a\,\diamondsuit\,b = \frac{a}{b}$ for all $(a,b)\in \mathbb R_{>0}\times \mathbb R_{>0}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(a,b,c)$ be the assertion that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ for positive real numbers $a,b$ and $c$. We start off with some minor observations. Claim : The binary operation $\diamondsuit$ must satisfy $a\diamondsuit 1 = a$. Proof : Say there exists $n \ne m$ such that $a \diamondsuit m = a \diamondsuit n$ for some $a \in \mathbb{R}_{>0}$. Then, comparing $P(a,a,n)$ and $P(a,a,m)$ we have, \[(a\diamondsuit a)m = a\diamondsuit (a \diamondsuit m) = a \diamondsuit (a \diamondsuit n) = (a \diamondsuit a)n\]which is a clear contradiction. Thus, there exist no such $n,m$ and the binary operation $\diamondsuit$ is 'injective in the second input'. Now note that from $P(1,a,1)$ we have \[1\diamondsuit (a \diamondsuit 1) = 1\diamondsuit a\]which due to our above observation implies $a \diamondsuit 1 =a$ as desired. To proceed, we define the function $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that $f(a)=1 \diamondsuit a$ for all $a \in \mathbb{R}_{>0}$. Now we note some properties of this function. Claim : For all positive reals $a$ and $b$, we have $a \diamondsuit b = af(b)$. Proof : First note that from $P(1,1,a)$ we have \[f(f(a))=1\diamondsuit f(a) = 1\diamondsuit(1 \diamondsuit a)= (1 \diamondsuit 1)a = a\]which implies that $f$ is an involution. Now, $P(a,1,f(b))$ yeilds, \begin{align*} a \diamondsuit (1 \diamondsuit f(b)) &= (a \diamondsuit 1) f(b) \\ a \diamondsuit f(f(b)) &= af(b)\\ a \diamondsuit b &= af(b) \end{align*}as claimed. Note that $f(1)=1$ since $1 \diamondsuit 1=1$. We further note that $P(a,b,c)$ gives \begin{align*} a \diamondsuit (b \diamondsuit f(c)) &= (a \diamondsuit b) f(c)\\ a \diamondsuit bf(f(c)) &= af(b)f(c)\\ af(bf(f(c))) &= af(b)f(c)\\ f(bc) &= f(b)f(c) \end{align*}for all pairs of positive reals $b$ and $c$. Thus, $f$ is multiplicative. Now, consider $0<a\le 1$. Define the function $g: \mathbb{R}_{>0} \to \mathbb{R}$ such that $g(x)=log_{a}(f(a^x))$. Note that, \begin{align*} f(a^{x+y}) &= f(a^x)f(a^y)\\ log_a(f(a^{x+y})) &= log_a(f(a^x)) + log_a(f(a^y))\\ g(x+y) &= g(x) + g(y) \end{align*}Hence, $g$ is Cauchy-additive. Further note that for all $a\le 1$, \[f(a) = \frac{1}{f\left(\frac{1}{a}\right)} \le \frac{1}{a}\]as a result of the second condition. But then for all $a\le 1$ and $x>0$, \begin{align*} f(a^x) & \le \frac{1}{a^x}\\ log_a(f(a^x)) & \le log_a\left(\frac{1}{a^x}\right)\\ g(x) & \le -x <0 \end{align*}Thus, the function $g$ is bounded above in the range $(0,\infty)$. Since we previously noted that it is also additive, we conclude that $g$ is linear over the positive reals. Thus, we let $g(x)=kx+c$ for constants $k$ and $c$ for all $x \in \mathbb{R}_{>0}$. Plugging this form back into the relation, \begin{align*} f(a^{x+y}) &= f(a^x)f(a^y)\\ a^{k(x+y)+c} &= a^{kx+c} \cdot a^{ky+c}\\ k(x+y) +c &= k(x+y) + 2c \end{align*}which implies that $c=0$. Further, since $f$ is an involution, \[a=f(f(a))=a^{k^{2}}\]Thus, $k\in \{1,-1\}$. Thus, $f(a)=a$ for all $a\le 1$ or $f(a)=\frac{1}{a}$ for all $a \le 1$. Since $f$ is multiplicative it follows from \[f(a)f\left(\frac{1}{a}\right)=f(1)=1\]that $f(a)=a$ for all $a \in \mathbb{R}_{>0}$ or $f(a) = \frac{1}{a}$ for all $a \in \mathbb{R}_{>0}$ as well, which finishes the proof.