socrates wrote: Find all real numbers $a$ for which there exists a non-constant function $f :\Bbb R \to \Bbb R$ satisfying the following two equations for all $x\in \Bbb R:$ i) $f(ax) = a^2f(x)$ and ii) $f(f(x)) = a f(x).$ Let $A=f(\mathbb R)$ Since non contant, $\exists u\in A$ such that $u\ne 0$ ii implies $f(u)=au$ and so : 1) $f(f(u))=af(u)=a^2u$ (using ii) ) 2) $f(f(u))=f(au)=a^2f(u)=a^3u$ (using i) ) So $a^3=a^2$ and $\boxed{a\in\{0,1\}}$ and it is trivial to check that both values fit.

Hello!My solution: If $a=0$ a function which satisfies is $f(x)=\left\{\begin{matrix} 0,x\geq 0\\ 2016,x<0 \end{matrix}\right.$ If $a=1$ one function which obviously satisfies is $f(x)=x$. Suppose now that $a \neq 0,1$.Setting $x \rightarrow \frac {f(x)} {a}$ in the first relation we have $a^2f(\frac {f(x)} {a})=f(f(x))=af(x) \Rightarrow f(\frac {f(x)} {a})=\frac {f(x)} {a}$ (1) But ii) with $a \neq 1$ implies that either $f$ has no fixed points or that its only fixed point is $0$. In the first case (1) gives contradiction.In the second (1) implies that $\frac {f(x)} {a}=0 \Rightarrow f(x)=0$ which is a constant function,contradiction. So the only solutions are $a=0,1$.

We claim that $a=0$ or $a=1$. For $a=0$ choose \[ f(x) = \begin{cases} 0 \text{ for all } x \in \mathbb{R} \setminus \{1 \} \\ 2 \text{ for } x=1. \end{cases} \]For $a=1$ choose $f(x)=x$. It is easy to check that both work. Now assume $a \neq 0$ and $a \neq 1$. It suffices to show that no other values of $a$ satisfy the FEs. Note that \[ af(f(x)) = f(f(f(x))) = f(af(x)) = a^2f(f(x)) \]So $f(f(x))=0$. That implies \[ 0 = f(f(x)) = af(x) \]which then implies $f(x)=0$. Contradiction, $f(x)$ is a non-constant function.

Let $P(x)$ be the assertion $f(ax)=a^2f(x)$ and let $Q(x)$ be the assertion $f(f(x))=af(x)$. Assume $a\ne0$. $P\left(\frac{f(x)}a\right)-Q(x)\Rightarrow f\left(\frac{f(x)}a\right)=\frac{f(x)}a$ $Q\left(\frac{f(x)}a\right)\Rightarrow f(x)=af(x)\Rightarrow a=1$ since $f$ is nonconstant. Now we check $a\in\{0,1\}$. If $a=0$, then $P(x)\wedge Q(x)$ iff $f(f(x))=0$, and the function $f(x)=\begin{cases}0&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ works. If $a=1$, then any function with $f(f(x))=f(x)$ satisfies, so $f(x)=x$ works. Hence the answer $\boxed{0,1}$.