= Solution
Since the <two-transitive group action> is primitive, the orbits of any nontrivial <normal subgroup> form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=NG_\alpha$, $N\cap G_\alpha=1$, and <conjugation> by $G_\alpha$ gives <automorphisms> of $N$. The <two-transitive group action> makes these <automorphisms> transitive on $N\setminus\{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. <Cauchy's theorem for finite groups> now shows $|N|=p^d$.
The <center of a group> of prime-power order is nontrivial and is a <characteristic subgroup>. The transitive <conjugation> action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an <elementary abelian group>, which we identify with the additive <vector space> $\mathbb F_p^d$. <Conjugation> by $G_\alpha$ is linear on this <vector space>. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. Consequently
$$
\boxed{n=p^d,\qquad G=N\rtimes G_\alpha\le AGL_d(p).}
$$
Here the <general affine group> acts by translations followed by invertible <linear maps>.
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