= Solution
Fix $\alpha\in\Omega$. If $\beta=g\alpha$, then normality of $H$ gives
$$
g(H\alpha)=(gHg^{-1})(g\alpha)=H\beta.
$$
Thus the transitive <group action> of $G$ permutes the <orbits of a group action> of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega|$. The only possibilities are $d=1$ and $d=|\Omega|$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega|$ and
$$
\boxed{H\text{ acts transitively on }\Omega}.
$$
Now suppose $H\cap G_\alpha=\{1\}$. Define the <orbit map>
$$
\theta:H\longrightarrow\Omega,
\qquad
\theta(h)=h\alpha.
$$
It is surjective by transitivity of $H$. If $\theta(h_1)=\theta(h_2)$, then $h_2^{-1}h_1\in H\cap G_\alpha$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_\alpha$,
$$
\theta(khk^{-1})
=khk^{-1}\alpha
=kh\alpha
=k\theta(h).
$$
\b[Thus $\theta$ is an <equivariant map>: <conjugation> by $G_\alpha$ on $H$ corresponds exactly to the given action of $G_\alpha$ on $\Omega$.]
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