Undergraduates at my university showed me this problem, which I found intriguing and now want to see the solution of:
Find all functions $f, g, h : \mathbb R_{> 0} \to \mathbb R_{> 0} $ such that
$\begin{cases}f \circ g = x^3 \\ g \circ h = x^4 \\ h \circ f = x^5\end{cases}.$
If we first assume that the functions take the form $x\mapsto x^r$ for $r\in \mathbb R$, we have a system
$\begin{cases}\deg(f) \deg(g) = 3 \\ \deg(g) \deg(h) = 4 \\ \deg(h) \deg(f) = 5\end{cases}$
with solution
$$\begin{cases}\deg f &= \frac{\pm\sqrt{15}}2 \\ \deg g &= \pm 2 \sqrt{3/5} \\ \deg h &= \pm 2 \sqrt{5/3}\end{cases}.$$
However, this does not mean these are the only functions. Is there a way to argue that solutions can be only of the form of this ansatz?
This answers the case where $f, g, h: {\mathbb R}\to{\mathbb R}$, which was the original question, it seems to have been modified since then.
From $f\circ g = x^3$ we get that $f$ is onto because $x^3$ is onto. From $h \circ f = x^5$ we get that $f$ is one-to-one because $x^5$ is one-to-one. Hence $f$ is bijective, hence \begin{equation} g = f^{-1}\circ x^3\qquad h = x^5\circ f^{-1} \end{equation} It follows that both $g$ and $h$ are bijective, but then $x^4 = g\circ h$ should be bijective, but this is false because $x^4$ is not onto.
Conclusion: there are no such functions.