= Solution
Choose $\alpha\in\Omega$ and identify $\Omega$ with $N$ by $n\mapsto n\alpha$. Normality means that the <point stabilizer> $H_\alpha$ acts on $N$ by conjugation, and
$$
h(n\alpha)=(hnh^{-1})\alpha\qquad(h\in H_\alpha).
$$
Double transitivity therefore says that $H_\alpha$ is transitive on $N\setminus\{1\}$. All nonidentity elements of $N$ have the same order. By <Cauchy's theorem for finite groups>, some element has order a prime $p$ dividing $|N|$, so that common order is $p$. Cauchy's theorem excludes every other prime from $|N|$, hence $N$ is a finite p-group.
By the center argument from Q1, $Z(N)\ne1$. The center is characteristic in $N$ and therefore invariant under the conjugation action of $H_\alpha$. Its nonidentity elements form a nonempty invariant subset of the transitive set $N\setminus\{1\}$, so $Z(N)=N$. Thus $N$ is abelian of exponent $p$:
$$
\boxed{N\cong(\mathbb F_p^d,+)\quad\text{for some }d\geq1.}
$$
This is <elementary abelian regular kernels in doubly transitive groups>; knowing only that all element orders agree would not by itself prove commutativity.
Every $h\in H$ is uniquely $nh_0$ with $n\in N$ and $h_0\in H_\alpha$, because regularity supplies the unique $n$ sending $\alpha$ to $h\alpha$. Hence $H=N\rtimes H_\alpha$. The conjugation action of $H_\alpha$ on $N$ is faithful: an element centralizing $N$ and fixing $\alpha$ fixes every $n\alpha$, and is the identity <permutation>. <Group automorphisms> of an elementary abelian p-group are exactly invertible $\mathbb F_p$-linear maps, so $H_\alpha\leq GL(d,p)$. Choosing a vector-space <basis> therefore gives
$$
\boxed{H\hookrightarrow\mathbb F_p^d\rtimes GL(d,p)=AGL(d,p).}
$$
Under this embedding, $N$ is the translation <subgroup> and the <point stabilizer> is a linear <subgroup> transitive on nonzero <vectors>. The action is assumed to have at least two points, as usual for a doubly transitive <group>.
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