Let $A, B:G\to \mathrm{Aut}(X)$ be two free group actions, where $X$ is a set and $G$ a group. A group action, $A$, is free if for all $g\in G\setminus{e}$, $A(g)x\neq x$ for all $x\in X$.Then there exists a function $f:X\to X$ such that for all $g\in G, x\in X$ $$f(A(g)x)= B(g)f(x).$$
I've been completely stumped for some time! How ought one think about such a question? I can't even think of a candidate $f$ or of some simple properties it ought satisfy! I've managed to dig up that the action being free is equivalent to $(g, x)\mapsto (A(g)x, x)$, $G\times X \to X\times X$ being an injection, but haven't been able to make use of this fact.
Hint: Write $X$ as union of orbits for the actions defined by $A$ and $B$, these orbits are isomorphic and can be identified to $G$ endowed with the unique orbit of the action $C:G\times G\rightarrow G$ defined by $C(g)g'=gg'$ since the action is free, you just have to find a bijection between the orbits of $A$ and the orbits of $B$.