 Solution

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,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 exactly two orbits on $\Omega \times \Omega$.