= Free finite actions with equal cardinality are isomorphic
{title2=$f(g\cdot s)=g*f(s)$}
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 $f(g\cdot s)=g*f(s)$ 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 $s_1,\ldots,s_r$ and $t_1,\ldots,t_r$, and define
$$
f(g\cdot s_i)=g*t_i.
$$
Every 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 $g*t_i\mapsto g\cdot s_i$ defines its inverse. Finally
$$
f(h\cdot(g\cdot s_i))=(hg)*t_i=h*(g*t_i)=h*f(g\cdot s_i),
$$
proving 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>.
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