firstly I apologize for my naive knowledge in group theory.
Let $G$ be a finite $p$-group with automorphism group ${\rm Aut}(G)$ and let $N$ be a maximal subgroup of $G$. Let $g$ be an element of G with $G = \langle g, N\rangle $ and let $\rho$ denote the automorphism of $N$ induced by conjugation by $g$. If $u$ is an element of $N$, let $\mu_u$ denote conjugation in $N$ by $u$. Then there is a bijection between \begin{eqnarray*} &&A_\alpha:=\{\phi\in {\rm Aut}(G)\ |\ \phi{|}_N=\alpha, \phi{|}_{G/N}=1_{G/N} \},\\ &&N_\alpha:=\{u\in N\ |\ [\rho,\alpha]=\mu_u, (g^p)^{\alpha}=(gu)^p\} \end{eqnarray*} given by $\phi\longrightarrow g^{-1}g^{\phi}$.
Please give me more details about the following fact:
If $\alpha$ has order a power of $p$ so does any element of $A_\alpha$.
Notation: By the symbols $g^{\phi}$ and $\phi{|}_N$.
I mean the image of $g$ under $\phi$ and restriction of $\phi$ to $N$, respectively.
Also $[\rho,\alpha]=\rho^{-1}\alpha^{-1}\rho\alpha$ denotes
commutator of $\rho$ and $\alpha$.
Thank you very much.
Edit:Since $|G/N|=p$, then any element of $G$ is of the form $g^in$, where $1\leq i\leq p-1$ and $n\in N$. How could we show that $(g^in)^{{\phi }^{p^t}}=g^in$ for some $t\geq 1$.