The group of automorphisms of a group $G$ that act trivially on $G/N$ $\textbf{Definition.}$ Let $G$ be a finite group which has only a minimal normal subgroup, say $N$. We say that an automorphism $\gamma:G \to G$ acts trivially on $G/N$ if the induced homomorphism $\gamma':G/N \to G/N$( $\gamma':Ng \mapsto Ng^ {\gamma}$) is the identity, that is, $g^{\gamma} \in Ng,~\forall g \in G$.
Let $\Gamma$ denote the group of those automorphisms of L that act trivially on $G/N$.
$\textbf{Statement:}$ If $N$ is abelian and complemented in $G$, then $|\Gamma|=(q-1)a$, where $q=|\mathrm{End}_G(N)|$ and $a$ is the number of complements for $N$ in $G$.
I'm trying to build an equivalence relation on the $\Gamma$ group, but I'm not getting an idea of how to proceed.
 A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\norm}{\trianglelefteq}$$\renewcommand{\phi}{\varphi}$$\DeclareMathOperator{\End}{End}$The following treatment is a bit sketchy, but hopefully correct, and comprehensible.
Suppose $\phi \in \End_{G}(N)$, that is, $\phi$ is an endomorphism of $N$ such that $(g^{-1} n g)^{\phi} = g^{-1} n^{\phi} g$, for $n \in N$ and $g \in G$. (In the following I will write $x^{g}$ for $g^{-1} x g$. Since $N^{\phi} \le N$, and $N^{\phi} \norm G$, if $\phi \ne 0$ it follows $\phi$ must be an automorphism of $N$. (Here the $0$ endomorphism is the one mapping the whole of $N$ onto $1$.)
If $\gamma \in \Gamma$, the restriction of $\gamma$ to $N$ is an element of $\End_{G}(N)$, as
$$
(g^{-1} n g)^{\phi}
=
(g^{\phi})^{-1} n^{\phi} g^{\phi}
=
g^{-1} m^{-1} n^{\phi} m g
=
g^{-1} n^{\phi} g,
$$
where $g^{\phi} = m g$ for some $m \in N$, and we have exploited the fact that $N$ is abelian.
Moreover $\gamma$ clearly sends a complement of $N$ to another complement of $N$.
Now fix a complement $H$ of $N$ in $G$. Let $K$ be another complement. Then $$
K
=
\Set{h h^{\psi} : h \in H},
$$
for some function $\psi :H \to N$. We have
$$
h_{1} h_{1}^{\psi} h_{2} h_{2}^{\psi}
=
h_{1} h_{2} (h_{1}^{\psi})^{h_{2}} h_{2}^{\psi},
$$
so that
$$\tag{cocycle}(h_{1} h_{2})^{\psi} = (h_{1}^{\psi})^{h_{2}} h_{2}^{\psi}.$$
Conversely any such cocycle $\psi$ defines a complement of $N$ in $G$, so that there are as many such cocycles $\psi$ as there are complements of $N$ in $G$.
Now for any of the $\theta \in \End_{G}(M)$, and any of the $a$ cocycles $\psi : H \to N$, define for $h \in H$ and $n \in N$
$$
(h n)^{\gamma}
=
h h^{\psi} n^{\theta}.
$$
Then
$$
(h_{1} n_{1} h_{2} n_{2})^{\gamma}
=
(h_{1} h_{2} n_{1}^{h_{2}} n_{2})^{\gamma}
=
h_{1} h_{2} (h_{1}^{\psi})^{h_{2}} h_{2}^{\psi}
(n_{1}^{\theta})^{h_{2}} n_{2}^{\theta}
$$
and
$$
(h_{1} n_{1})^{\gamma} (h_{2} n_{2})^{\gamma}
=
h_{1} h_{1}^{\psi} n_{1}^{\theta} 
h_{2} h_{2}^{\psi} n_{2}^{\theta} 
=
h_{1} h_{2}
(h_{1}^{\psi})^{h_{2}} (n_{1}^{\theta})^{h_{2}} 
h_{2}^{\psi} n_{2}^{\theta},
$$
which equals the previous expression, as $N$ is abelian. Therefore $\gamma \in \Gamma$.
