On Hungerford lemma about normalizer of $p$-subgroup

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This might be a simple question, but I would like to ask for clarification on Lemma 5.5 in Hungerford chapter II.

It says: "If $H$ is a $p$-subgroup of a finite group $G$, then $[N_G(H):H]\equiv [G:H] \text{ (mod } p)$."

It uses the previous Lemma 5.1: "If a finite $p$-group $H$ acts on a finite set $S$ and if $S_0=\{x\in S\mid hx=x \text{ for all } h\in H\}$, then $|S|\equiv|S_0| \text{ (mod } p)$".

So letting $S$ be the set of all left cosets and $H$ acting on it by the usual left translation, then $|S|$ is the index of $H$ in $G$.

But for $S_0$, it says ($xH\in S_0$) iff ...etc... iff ($x^{-1}hx\in H$ for all $h\in H$) iff* ($x^{-1}Hx=H$).

I do not see why the iff* above holds, especially from left to right. Hopefully I can get some guidance. Thank you very much.

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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$This is the left-to-right implication.

Your assumption is that $x H \in S_{0}$. This is is equivalent to $h x H = x H$ for all $h \in H$, which can be rewritten, multiplying on the left by $x^{-1}$, as $x^{-1} h x H = H$ for all $h \in H$, which is equivalent to $x^{-1} h x \in H$ for all $h \in H$, which is in turn equivalent to $x^{-1} H x \subseteq H$.

Since $H$ is finite, and $\Size{x^{-1} H x} = \Size{H}$, you get $x^{-1} H x = H$.