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Let $G$ be a finite group which operates on two finite sets $E_1$ and $E_2$. Say that $E_1$ and $E_2$ are weakly $G$-isomorphic if for every $g \in G$, $\mathrm{Card}(E_1^g)=\mathrm{Card}(E_2^g)$, with $E^g$ the set of points fixed by $g$. An equivalent definition is : for every $g$ there exists a bijection $f_g : E_1 \to E_2$ such that $f_g(gx_1)=gf_g(x_1)$ (for every $x_1 \in E_1$) (this is not obvious but not difficult either).

I would like to prove that if $H_1,H_2$ are subgroups of $G$, then the $G$-sets $G/H_1$ and $G/H_2$ are weakly $G$-isomorphic iif for every conjugacy class $C$ in $G$, $\mathrm{Card}(C \cap H_1) = \mathrm{Card}(C \cap H_2)$.

Now for $xH \in G/H$ to be a fixed point of $g$, a necessary and sufficient condition is that $x^{-1}gx \in H$. We can't define a mapping $xH \mapsto x^{-1}gx \in C \cap H$, because if $xH=yH$ there is no reason for $x^{-1}gx$ to be the same as $y^{-1}gy$. But we can map the fixed point $xH$ to the set of $h^{-1}x^{-1}gxh$, i.e. the conjugacy class of $x^{-1}gx$ for $H$ (note that $H$ operates on $C \cap H$ by conjugation).

I've been playing around with this kind of things for quite some time but eventually didn't succeed in construction a bijection from $C \cap H_1$ onto $C \cap H_2$, assuming $G/H_1$ and $G/H_2$ are weakly $G$-isomorphic.

Does anyone have an idea?

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