 Solution

By 3(a), identify $\Omega$ with $V={\mathbb{F}}_p^d$ and let $H=G_0\le G{L}_d(p)$. 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 $d\ge 2$, the pairs $(u,au)$ with $a\ne 0,1$ 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 $p\le 3$, incompatible with four-transitivity, whose degree is at least four.

Hence $p=2$. If $d\ge 3$, 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 $AG{L}_2(2)$ has order $4(4-1)(4-2)=24$. Four-transitivity on four points requires $24\mid |G|$, and $G\le AG{L}_2(2)\le S_4$. Consequently