Suppose that the cyclic group $G$ acts on a set $S$ and $g_1$ and $g_2$ generate $G$. Show that $|$Fix $g_1|=|$Fix $g_2|$. We know that there is a function $\psi: G\times S\rightarrow S$ with the properties (if $\psi(g,s)$ is written as $g(s)$) that $(g^\star\cdot g^*)(s)=g^\star(g^*(s))$ and $e_G(s)=s~\forall g^\star,g^*\in G, s\in S$.
I am just looking for a hint to start with.
Hint:
If $g$ generates $G$, then the elements fixed by $g$ under the action are exactly the elements fixed by $G$ under the action.