 Solution

For a free group action, every stabilizer subgroup is $\{e\}$. By the orbit-stabilizer theorem, every orbit of a group action has cardinality $|G|$. Hence the two free group actions have the same numberof orbits of a group action. Choose one representative $s_i$ from each $\cdot$ orbit of a group action and one representative $t_i$ from each $\ast$ orbit of a group action, and match them using $1\le i\le r$.

Every $s\in S$ has a unique expression $s=g\cdot s_i$. The orbit of a group action determines $i$, and the free group action property determines $g$: if $g\cdot s_i=h\cdot s_i$, then $(h^{-1}g)\cdot s_i=s_i$, forcing $g=h$. Therefore the ruledefines a function $f:S\to S$ without ambiguity. For $h\in G$,which proves the required identity for every $s$.

Indeed, $f$ is a bijection: its inverse is $g\ast t_i\mapsto g\cdot s_i$, well defined because the second is also a free group action. If $S$ is empty, the unique empty function already satisfies the identity. Thus the result holds in every case, and proves the stronger statement that free finite actions with equal cardinality are isomorphic.