= Solution
{c}
The hypothesis is a <two-transitive group action>. Given $\alpha,\beta\ne\omega_1$, apply it to $(\omega_1,\alpha)$ and $(\omega_1,\beta)$. The resulting element fixes $\omega_1$ and sends $\alpha$ to $\beta$, so the <stabilizer subgroup> acts transitively on the remaining $n-1$ points.
The action on all of $\Omega$ is also transitive. By the <orbit-stabilizer theorem>,
$$
|G|=n|\operatorname{Stab}_G(\omega_1)|,\qquad
|\operatorname{Stab}_G(\omega_1)|=(n-1)|\operatorname{Stab}_G(\omega_1,\omega_2)|.
$$
Thus $\boxed{n(n-1)\mid |G|}$. The diagonal pairs form one orbit, while two-transitivity makes all ordered unequal pairs a second orbit. Equality is preserved by a group action, so these cannot mix. There are \b[exactly two orbits on $\Omega\times\Omega$].
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