Let $p \neq 2$ be a prime, $\Sigma_p$ be the symmetric group on $p$ elements, and $Z_{(p)}$ be the integers localized at $p$.
Let $G$ be a finite group with Sylow $p$-subgroup $\mathbb{Z}/p$ and $\Sigma_p \subset G$.
Then we have
\begin{equation}\tag{1} H^*(G,\mathbb{Z}_{(p)}) = H^*(\mathbb{Z}/p,\mathbb{Z}_{(p)})^W \end{equation}
where $W\subset \text{Aut}(\mathbb{Z}/p)$ is the image of the normalizer $N(\mathbb{Z}/p) \subset G$ in $\text{Aut}(\mathbb{Z}/p)$ under the natural action of $N(\mathbb{Z}/p)$ on $\mathbb{Z}/p$ (The superscript denotes invariants here, I think, in the sense that it is the submodule that's unchanged under the action of $W$).
We have a theorem that if $G$ is a finite group with $H$ a $p$-Sylow subgroup, then $\text{res}^G_H$ maps $H^n(G,M)_{(p)}$ isomorphically onto the set of $G$-invariant elements of $H^n(H,M)$, where $H^n(G,M)_{(p)}$ is the $p$-primary component of $H^n(G,M)$. Here the invariants are defined as follows: Let $G$ be an arbitrary group and $H \subset G$. An element $z \in H^*(H,M)$ is $G$-invariant if $\text{res}^H_{H\subset gHg^{-1}}z = \text{res}^{gHg^{-1}}_{H \cap gHg^{-1}} gz$ for all $g\in G$. It is known that if $z = \text{res}^G_H w$ for some $w\in H^*(H,M)$, then $z$ is $G$-invariant. Here res is the standard restriction map in group cohomology, $\text{res}^G_H: H^*(G,M) \to H^*(H,M)$.
My questions concerning this isomorphism are:
Does the "natural action" of $N(\mathbb{Z}/p)$ on $\mathbb{Z}/p$ refer to conjugation by the element in the normalizer?
I don't understand why $H^*(\mathbb{Z}/p,\mathbb{Z}_{(p)})^W$ is the $G$-invariant elements. If that's the case, then we're done here.
Define the holomorph $H = \mathbb{Z}/p \rtimes \text{Aut}(\mathbb{Z}/p)$ (with the identity map in the semidirect product); it is known that we can embed $H$ in $\Sigma_p$. Since $H \subset \Sigma_p\subset G$, then the restriction map \begin{equation}\tag{2} H^*(G; \mathbb{Z}_{(p)}) \to H^*(\Sigma_p;\mathbb{Z}_{(p)}) \end{equation} is an isomorphism. Here I don't understand why we get the isomorphism above. Could someone explain?