= Solution
By 3(a), identify $\Omega$ with $V=\mathbb F_p^d$ and let $H=G_0\le GL_d(p)$. Four-transitivity means $H$ is transitive on ordered triples of distinct nonzero <vectors>, and in particular on ordered pairs of distinct nonzero <vectors>.
If $p>2$ and $d\ge2$, the pairs $(u,au)$ with $a\ne0,1$ and $(u,v)$ with $u,v$ <linearly independent> cannot be interchanged by a <linear map>. Thus $d=1$ if $p>2$. Then $|H|\le p-1$, whereas pair-transitivity on the $p-1$ nonzero points requires $(p-1)(p-2)\le |H|$. This forces $p\le3$, incompatible with four-transitivity, whose degree is at least four.
Hence $p=2$. If $d\ge3$, the triples $(u,v,u+v)$ and $(u,v,w)$ for <linearly independent> $u,v,w$ cannot be interchanged: the first is <linearly dependent> and the second is not. Thus $d=2$. The <general affine group> $AGL_2(2)$ has order $4(4-1)(4-2)=24$. Four-transitivity on four points requires $24\mid |G|$, and $G\le AGL_2(2)\le S_4$. Consequently
$$
\boxed{n=4,\qquad G=AGL_2(2)\cong S_4.}
$$
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