The answer is $f(x)\equiv x, f(x)\equiv -x$ only. They both work. Let $P(x,y)$ denote the assertion that the multiset $\{(f(xf(y)+1),f(yf(x)-1)\}$ is identical to the multiset $\{xf(f(y))+1,yf(f(x))-1\}$. $P(0,0) \to \{f(1), f(-1)\}=\{1,-1\} \to f(f(1))=1, f(f(-1))=-1$ Claim 1: $f$ is surjective. Proof$P(x-1,1)$: $\{f((x-1)f(1)+1), f(f(x-1)-1)\} = \{x, f(f(x-1))-1\}$. It follows there exists $t$ such that $f(t)=x$ for all $x\in \mathbb{R}$ Claim 2: $f(0)=0$. ProofSuppose $f(0)\ne 0, f(a)=0 \rightarrow a\ne 0$. Then $P(a,a): \{f(af(a)+1), f(af(a)-1) \} = \{af(f(a))+1, af(f(a))-1\} $ We get $\{1,-1\}=\{af(f(a))+1, af(f(a))-1\}$ so $af(f(a))=0 \rightarrow f(0)=0$. Case 1: $f(1)=1$. I claim $f(x)\equiv x$ Assume for contradiction $f(x)\ne x$ for some $x\ne 0$. $P(x-1,1): \{f(x), f(f(x-1)-1) \} = \{ f((x-1)+1), f(f(x-1)-1) \} = \{ (x-1)(1)+1, f(f(x-1))-1\} = \{x,f(f(x-1))-1\}$ Since $f(x)\ne x$ it follows that $f(f(x-1)-1)=x, f(x)=f(f(x-1))-1$ (1) $P(1,1+x): \{f(f(x+1)+1), f(x) \} = \{ f(f(x+1)+1), f((x+1)-1) \} = \{ f(f(x+1))+1, (x+1)-1\} = \{ f(f(x+1))+1, x\}$ Since $f(x)\ne x$ it follows that $$f(f(x+1)+1) = x, f(x)=f(f(x+1))+1 (eq.3)$$ Claim 3: If $f(a)=0$ for some $a\ne 0$, (1) and (3) cannot simultaneously hold. Proof$P(a,y): \{f(af(y)+1), f(yf(a)-1) \} = \{af(f(y))+1, yf(f(a))-1\}$. By claim 2, $f(0)=0$ so $\{f(af(y)+1), f(-1) \} = \{af(f(y))+1, -1\}$ It follows that $f(af(y)+1)=af(f(y))+1$ for all reals $y$. Since $f(y)$ covers all reals, we can rewrite this as $f(ay+1)=af(y)+1$ for all reals $y$. Similarly, $P(y,a)$ gives $f(ay-1)=af(y)-1$ for all reals $y$. Therefore, $f(y+1)-f(y-1)=2$ for all reals $y$. $f(f(x+1)+1) = f(f(x-1)+2+1)=f(f(x-1)+4-1)=f(f(x-1)+2-1)+2=f(f(x-1)-1)+4$ By (1) RHS is $x+4$ and by (3) LHS is $x$, absurd! Therefore, $f(a)=0 \rightarrow a=0$. We can see if $f(f(a))=0 \to f(a)=0 \to a=0$ Claim 4: $f$ is injective. ProofAssume for contradiction $f(u)=f(v), u\ne v$. $P(u,y): \{f(uf(y)+1), f(yf(u)-1)\} = \{uf(f(y))+1, yf(f(u))-1\}$ $P(v,y): \{f(vf(y)+1), f(yf(v)-1)\} = \{vf(f(y))+1, yf(f(v))-1\}$ Notice $f(u)=f(v)$ so $f(yf(u)-1)=f(yf(v)-1)$ and $yf(f(u))-1=yf(f(v))-1$ Assume for contradiction $f(yf(u)-1)\ne yf(f(u))-1$ for some $y$, then $y\ne 0$ since $f(-1)=-1$. We have $f(yf(u)-1)=uf(f(y))+1$ and $f(yf(v)-1)=vf(f(y))+1$ so $f(f(y))=0$, contradicting our lemma. Therefore, $f(yf(u)-1)=yf(f(u))-1$ for all reals $y$. Similarly, $f(yf(u)+1)=yf(f(u))+1$ for all reals $y$, so (1) and (3) cannot hold by repeating that same argument. Finish: Main idea is to consider the number $f(x+1)+1=f(x-1)-1$. What happens if it were not a fixed point? finishNow with this in mind, we know $x= f( f(x+1)+1) = f( f(x-1)-1)$, so $f(x+1)+1=f(x-1)-1$(eq A) We also know $f(x)=f(f(x-1))-1=f(f(x+1))+1$ (eq B) Let $m=f(x-1)-1=f(x+1)+1$. If $f(m)\ne m$, we have $f(m+1)+1=f(m-1)-1$. Therefore, $f(f(x-1))+1=f(f(x+1))-1$, but Eq B says $f(f(x+1))+1=f(f(x-1))-1$, contradiction. Therefore, $f(m)=m$. We also know $f(f(x+1)+1)=x$, so $m=f(m)=x$, contradicting our assumption $f(x)\ne x$. The case where $f(1)=-1, f(-1)=1$ is similar and uses almost the same way to prove $f(x)\equiv -x$.

First note that there is such a $c$. $S(c,c)$ gives $0=f(1)+f(-1)=2cf(0)$ which in any case gives $f(0)=0$ .Now suppose that $c\neq 0$ $M(x,c)$ gives $\{ f(1),f(cf(x)-1)\}=\{1,cf(f(x))-1\}$ ,since $f$ is surjective this forces that $f(1)=1$ and $f(cf(x)-1)=cf(f(x))-1$ which because $f(x)$ takes all real values gives $f(cx-1)=cf(x)-1$ $M(c,x)$ similarly gives $f(cx+1)=cf(x)+1$ combining these we get $f(x+1)-f(x-1)=2$ Now $M(c-1,1)$ gives $\{0,f(f(c-1)-1)\}=\{c,f(f(c-1))-1\}$ and as $c\neq 0$ we get $f(f(c-1))=1$ Similarly $M(1,c+1)$ gives $\{f(f(c+1)+1),0\}=\{f(f(c+1))+1,c\}$ and we get $f(f(c+1))=-1$ But this is clearly a contradiction since $f(f(c+1))=f(f(c-1)+2)=f(f(c-1))+2$

Suppose that $f(a)=f(b)\neq 0$. We should have that $f(f(a))=f(a)$ or else use claim 2 to get that $a=b$. $M(x,a)$: $\{ f(xf(a)+1),f(af(x)-1)\}=\{xf(a)+1,af(f(x))-1\}$ $M(x,b)$: $\{ f(xf(a)+1),f(bf(x)-1)\}=\{xf(a)+1,bf(f(x))-1\}$ so either $xf(a)+1$ is a fixed point or $f(xf(a)+1)=af(f(x))-1=bf(f(x))-1\Rightarrow f(f(x))=0\Rightarrow x=0$. So in particular $-1$ is a fixed point which is a contradiction.

$S(1,-1)$ gives that $f(2)=-2$ .Suppose there is a fixed point $z$ then $\frac{1}{z}$ is also a fixed point and $S(z,\frac{1}{z})$ gives $f(2)=2$

$S(x,-1): f(x+1)+f(-f(x)-1)=-x-f(f(x))$ $S(f(x),-1): f(f(x)+1)+f(-f(f(x))-1)=-f(x)+x$ $S(f(f(x)),-1): f(f(f(x))+1)+f(x-1)=-f(f(x))+f(x)$ add them.

$S(\frac{1}{f(x)},-f(f(x)))$ gives:$f(\frac{x}{f(x)}+1)+f(-2)=-\frac{f^4(x)}{f(x)}-\frac{f(f(x)}{f^3(x)}=\frac{f(f(x))}{x}+1$ $S(-f(f(x)),\frac{1}{f(x)})$ gives: $f(0)+f(\frac{x}{f(x)}-1)=-\frac{f(f(x))}{f^3(x)}-\frac{f^4(x)}{f(x)}=\frac{f(f(x))}{x}+1$ add these to get $2\frac{f(f(x))}{x}=f(\frac{x}{f(x)}+1)+f(\frac{x}{f(x)}-1)=-2f(f(\frac{x}{f(x)}))\Rightarrow f(f(\frac{x}{f(x)}))=-\frac{f(f(x))}{x}$ apply this relation to $x=:f(f(x))$ to get $f(f(-\frac{f(f(x))}{x}))=-\frac{f^4(x)}{f(f(x))}=\frac{f(x)}{f(f(x))}$ thus $\frac{f(x)}{f(f(x))}=f^4(\frac{x}{f(x)})=-f(\frac{x}{f(x)})$

$M(-\frac{2}{f(x)},x)$ gives: $\{1,f(xf(-\frac{2}{f(x)})-1)=\{-\frac{2f(f(x))}{f(x)}+1,xf(f(-\frac{2}{f(x)}))-1\}$ thus $xf(f(-\frac{2}{f(x)}))=2$ put $\frac{1}{x}$ instead of $x$: $f(f(-2f(x)))=2x$ and $f$- both sides to get the result.