I don't have question, I just want a verification of my proof attempt for the following question:
$G$ is any group which might be infinite, and $H$ is a subgroup of $G$, where $H$ is finite.
Define $L$ as the set of all left cosets of $H$ in $G$. Let $\phi_H=\{xH\in L \mid hxH=xH, \forall h\in H\}$ and $N=\{g\in G\mid gHg^{-1}=H\}.$
Prove: $|\phi_H|=(N:H).$
My attempt
If I can show $\frac{N}{H}=\phi_H$, then it automatically completes the proof.
(1) to show $\frac{N}{H}\subseteq \phi_H$
$\forall gH\in \frac{N}H$, since $g\in N$, we have $h(gH)=(hg)H=(gh')H=g(h'H)=gH$, for $\forall h\in H$.
Therefore, $gH\in \phi_H$
(2) to show $\phi_H\subseteq \frac{N}{H}$
$\forall xH\in \phi_H$, to show $xH\in \frac{N}{H}$, it is equivalent to show $x\in N$. Since $N$ is a subgroup of $G$, to show $x\in N$, it is equivalent to show $x^{-1}\in N$.
$\forall h\in H$, since $xH\in \phi_H$, then $\exists h_1, h_2\in H$, such that $hxh_1=xh_2$
$\Rightarrow x^{-1}hx=h_2 h_1^{-1}\Rightarrow x^{-1}hx\in H\Rightarrow x^{-1}H (x^{-1})^{-1}=H\Rightarrow x^{-1}\in N\Rightarrow x\in N$.
Therefore, $xH\in \frac{N}{H} \quad \Box$
If $H$ is finite, your proof is correct but could be much shorter: for any $x\in G,$ $$\begin{align}xH\in\phi_H&\iff\forall h\in H,~x^{-1}hxH=H\\&\iff\forall h\in H,~x^{-1}hx\in H\\&\iff x^{-1}Hx\subseteq H\\&\iff x^{-1}Hx=H\\&\iff H=xHx^{-1}\\&\iff x\in N.\end{align}.$$
If $H$ is infinite, your statement may be false because $gHg^{-1}\subseteq H$ does not imply $gHg^{-1}=H$. The flaw in your proof is when you claim that your $N$ is closed under $x\mapsto x^{-1}.$