= Solution
{c}
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 number
$$
r=\frac{|S|}{|G|}
$$
of <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 $*$ <orbit of a group action>, and match them using $1\leq i\leq r$.
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 rule
$$
\boxed{f(g\cdot s_i)=g*t_i}
$$
defines a <function> $f:S\to S$ without ambiguity. For $h\in G$,
$$
f\bigl(h\cdot(g\cdot s_i)\bigr)
=f\bigl((hg)\cdot s_i\bigr)
=(hg)*t_i
=h*(g*t_i)
=h*f(g\cdot s_i),
$$
which proves the required identity for every $s$.
Indeed, $f$ is a <bijection>: its inverse is $g*t_i\mapsto g\cdot s_i$, 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>.
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