(Similar to the above question, but unfortunately I could not see where to go with the above hint. I'm sure it's obvious, but right now I feel that I am missing the point - even just a hint giving a small explanation would be appreciated.)
We say that a group, $G$, acts freely on a set S if for any non-trivial $g \in G$, and any $s \in S$, we have $g · s = s$. Let $φ(g, s) = g · s$ and $ψ(g, s) = g ∗ s$ be two free actions of a finite group G on a finite set S. Show that there is a function $f : S → S$ such that $$f(g · s) = g ∗ f(s)$$ for all $g ∈ G, s ∈ S$.
I am able to prove this in the case that $G$ is cyclic, but not otherwise. I am thinking that a possible way forward might be to consider the orbits of $S$ as we can see (from the Orbit-Stabiliser theorem, for example) that each orbit of $S$ has the same size for both actions; at this point I am just confusing myself though, and I would appreciate any hints on how to proceed.
Let $I$ be the set of orbits of the action $\varphi$ of $G$ on $S$, and let $\{ \alpha_i : i \in I \}$ be a set of representatives of the se orbits. Since the action is free, for each $s \in S$, there exists a unique $i \in I$ and $h \in G$ such that $h \cdot \alpha_i = s$.
Now, choose any $\beta \in S$ and define $f:S \to S$ by $f(h \cdot \alpha_i) = h * \beta$. By the uniqueness property above, this defines $f(s)$ uniquely for each $s \in S$.
We have to check that $f(g \cdot s) = g * f(s)$ for all $g \in G$ and $s \in S$. By the above, we can write each $s \in S$ as $h \cdot \alpha_i$ for some $i \in I$ and $h \in G$. Then $$f(g \cdot s) = f(g.(h \cdot \alpha_i)) = f((gh) \cdot \alpha_i) = (gh) * \alpha = g*(h* \alpha) = g*f(h \cdot \alpha) =\hbox{$ g*f(s)$}.$$