Free finite actions with equal cardinality are isomorphic ($f(g\cdot s)=g\ast f(s)$)

Let a finite group $G$ have free group actions on finite sets $S,T$ of equal cardinality. There is a bijection $f:S\to T$ satisfying $f(g\cdot s)=g\ast f(s)$ for every $g\in G$, $s\in S$.

By the orbit-stabilizer theorem, every orbit of a group action in either set has cardinality $|G|$. Thus both have $r=|S|/|G|$ orbits of a group action. Choose representatives $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$, and defineEvery point has a unique displayed representation: its orbit of a group action determines $i$, and the free group action property makes $g$ unique. Thus $f$ is well defined, and $g\ast t_i\mapsto g\cdot s_i$ defines its inverse. Finallyproving the required compatibility. Empty sets are covered by the unique empty bijection. This shows that a finite free group action is determined up to such an isomorphism by its number of orbits of a group action.