Is there a nontrivial solution to $f(f(f(x)))=-8x$? Let $f:\mathbb R\to\mathbb R$ be a continuous function such that $f(f(f(x)))=-8x$. Must we have $f(x)=-2x$? I can prove this if I assume that $f$ is continuously differentiable everywhere, but is that condition necessary? I was inspired to ask this question after I saw the thread a continuous function satisfying $f(f(f(x)))=-x$ other than $f(x)=-x$ .
Some facts I've discovered, which you may use without proof:


*

*$f$ is bijective, decreasing and therefore $f$ is a homeomorphism and differentiable almost everywhere

*$f(0)=0$ because if $f(0)=a$, then $f(f(a))=0$, so $-8a=f(f(f(a)))=f(0)=a\implies a=0$.

*If $x\neq 0$, then the list $\ldots,f^{-2}(x),f^{-1}(x),x,f(x),f^2(x),\ldots$ has no duplicate members, where $f^n$ is an $n$th functional power of $f$.

*If $f$ is differentiable at $0$, then $f'(0)=-2$ because $f'(0)f'(0)f'(0)=-8$.

*$x$ and $f(x)$ have opposite signs if $x\neq 0$, because $f(0)=0$ and $f$ is decreasing.

*$f\circ f$ restricted to the positive numbers is an increasing function.

*If we assume $f$ is continuously differentiable everywhere, then $f(x)=-2x$. Proof: Let $a_n=f^n(x)$ for some $x\neq 0$. Then $f'(a_1)f'(a_2)f'(a_3)=-8$, and $f'(a_2)f'(a_3)f'(a_4)=-8$ which gives us $f'(a_k)=f'(a_{k+3})$ for all integers $k$. Since $a_{3k}=(-8)^kx\to 0$ when $k\to-\infty$, we get $f'(x)=\lim_{k\to-\infty}f'(f^{-3k}(x))=f'(0)=-2$ and therefore $f(x)=-2x$ is the only solution.
 A: I will set $f(-x)=-f(x)$ and $g(x)=-f(x)$.
Now $g(g(g(x)))=-f(-f(-f(x)))=-f(f(f(x)))=8x$.
So, I am searching a continuous $g$ from $\mathbb R^+$ to $\mathbb R^+$.
Now, I am regarding $h(x)=\log(g(\exp(x)))$ where $h(x)$ is a function from $\mathbb R$ to $\mathbb R$ that is continuous.
Then 
$$\begin{multline}
h(h(h(x)))=\log(g(\exp(\log(g(\exp(\log(g(\exp x))))))))\\ =\log(g(g(g(\exp(x))))=\log(8\exp(x))=\log 8 + x=x+3\log 2.\end{multline}$$
Now, set $i(x)=h(x\log 2)/\log(2)$. Clearly, $i(i(i(x)))=x +3$.
Now, it is easy to see that it is enough to find three continuous increasing bijections of intervals, so that their concatenation is the identity.
For example, let $a(x)=x$, $b(x)=x^2$, and $c(x)=\sqrt x$ be such continuous increasing bijections for the interval $(0,1)$.
Now, take
$$i(x)=\begin{cases}
a(\{x\})+\lfloor x\rfloor +1 \text{ if $3n \le \lfloor x \rfloor < 3n+1$ }\\
b(\{x\})+\lfloor x\rfloor +1 \text{ if $3n+1 \le \lfloor x \rfloor < 3n+2$ }\\
c(\{x\})+\lfloor x\rfloor +1 \text{ if $3n+2 \le \lfloor x \rfloor < 3n+3$}\\ \end{cases}$$
where $\{x\}$ denotes the fractional part of $x$.
$i(x)$ is certainly continuous, increasing and has the right concatenation property. 
This gives one of many continuous solutions for the original equation.

In case you are wondering why this is substantially different from $f(f(f(x))))=-x$, note that for these equations the important object is the structure of the graph of real numbers with arrows from $x$ to $g(x)$ where $g$ is the function on the right-hand-side.
For $-8x$, the real numbers are split in 0 and doubly infinite chains, while for $-x$, the real numbers are split in 0 and circles of length 2.
To go to $f$ from the infinite chains, one has to take three of the chains and cycle through them to have a chain of $f$. There are so many ways of doing that that it is quite hopeful that some of them are continuous.
If you try to combine three of the circles of length 2 in a similar way, this will not give a monotone function, however, $f$ is bijective and continuous, so it has to be monotone. So, the only possibility is that $f$ maps each circle to itself which gives the unique solution.
So, in summary, chains are easier to reconcile with monotony than circles.
A: First, I'll use the notation $ f \circ f \circ f $.
Note that $ \frac {d} {dx} f\circ f\circ f = (f' \circ f \circ f)(f'\circ f)f'=C$.
Denote $ A=(f' \circ f \circ f), B=(f'\circ f)$ you get that $ A,B,f'\ne 0$ everywhere.
 $ \frac {d^2} {dx^2} f\circ f\circ f = (f' \circ f \circ f)(f'\circ f)f'=(f''\circ f\circ f)Bf'+A(f''\circ f)f'\cdot f' + ABf''=0$.
Since $A,B,f'\ne 0$ this could hold everywhere iff $f''=0$ everywhere, and you get that the solution has to be a linear function.
From what you already achieved it's easy to see that the only such polynomial is $-2x$.
