Is $AGL(3,2)$ self-normalizing in $S_8$? Let $AGL(3,2)$ be a group of all affine permutations consisting of maps
$$
x \to Ax + b,
$$
where $x$ and $b$ are vectors of length $3$ over $GF(2)$ and $A$ is an invertible $3\times 3$ matrix over $GF(2)$. Using some numeration of vectors we can consider $AGL(3,2)$ as a subgroup of $S_8$. My experimets with computer algebra system Sage have shown that normaliser of $AGL(3,2)$ in $S_8$ is equal to $AGL(3,2)$. But I can not prove this. Can somebody help me with hints or full solution.
 A: First of all, computations with Sage is a full solution, in my opinion. A better phrase for what you're asking for is an easy proof.
Here is a proof that in general $\mathrm{AGL}(n, p)$ is self-normalizing in $\mathrm{Sym}(\mathbf{F}_p^n) \cong S_{p^n}$.
Let $V \le \mathrm{Sym}(\mathbf{F}_p^n)$ be the group of all $p^n$ translations, so $V \cong \mathbf{F}_p^n$. Then $G =\mathrm{AGL}(n, p)$ is the normalizer of $V$. On the other hand $G / V \cong \mathrm{GL}(n, p)$ and $\mathrm{GL}(n, p)$ has no normal $p$-subgroup, so $V$ is the largest normal $p$-subgroup of $G$ and therefore characteristic in $G$, so $N(G) \le N(V)$. Therefore $G \le N(G) \le N(V) = G$, so $G = N(G)$.
A: The affine subgroup $N$ of ${\rm AGL}(3,2)$, is elementary abelian of order $8$, and it is characteristic in ${\rm AGL}(3,2)$ because it is the largest normal $2$-subgroup.
Since $N$ is abelian and acts regularly, it is self-centralizing in $S_8$ (I'll leave you to prove that), so $N_{S_8}(N)/C_{S_8}(N)$ is isomorphic to a subgroup of ${\rm Aut}(N)$, which is isomorphic to ${\rm GL}(3,2)$ (because $N$ can be regarded as a $3$-dimensional vector space over ${\mathbb F}_2$).
So, since since the whole of ${\rm GL}(3,2)$ is induced by conjugation in ${\rm AGL}(3,2)$, we must have $N_{S_8}({\rm AGL}(3,2)) = N_{S_8}(N) = {\rm AGL}(3,2)$.
