$G$-set isomorphism equivalence Why are the following statements equivalent? (Let $P$ and $Q$ be groups and $\varphi$ an isomorphism between them. Let $X$ be a finite $P$-set):


*

*${}_PX\cong {}_P^{\varphi}X$ as $P$-sets.

*$\vert {}_PX^R\vert=\vert {}_P^{\varphi}X^R\vert$ for all $R\leq P$


Where ${}_PX$ is the set $X$ under the action of $P$, the set ${}_P^{\varphi}X$ is the set $X$ under the action of $P$ given by $p\ast x=\varphi(p)\cdot x$ for $p\in P$ and $x\in X$, and $X^R$ is the set of fixed points under the action of $R$.
 A: I believe this is using a basic fact about finite $G$-sets, which is that they are determined up to isomorphism by the number of fixed points for every subgroup of $G$.  
The proof is easy: Let $X$ and $X'$ have the same number of fixed points for every subgroup $H\subset G$.  First note that a subgroup $H\subset G$ has a fixed point in $X$ iff it is contained in the stabilizer of a point in $X$.  So if $H$ is a subgroup of minimal index such that $X^H\neq \emptyset$, then it must be exactly the stabilizer of a point $x\in X$, and the orbit $G\cdot x$ is isomorphic to the cosets $G/H$.  By the same argument, there exists $x'\in X'$ with the same property, so the orbit $G\cdot x'\cong G/H$.
So, now we use induction on the size: there's an isomorphism $G\cdot x\cong G\cdot x'$ sending $x\mapsto x'$.  Thus, to define an isomorphism between $X$ and $X'$ we just have to define on on the complements of these orbits, which is possible by induction.  Basically, you just repeat the same arguments until you've matched all the orbits.
