 Solution

Adjoin $\infty$ to $\Omega =\{1,\dots ,n\}$ and take $G=A_n$ fixing it. Choose three distinct points $a,b,c\in \Omega$ and put $t=(\infty a)(bc)$. This is an even involution interchanging $\infty$ and $a$. Conjugation by it preserves $H=G_a$, because it permutes the remaining $n-1$ points and preserves parity. Thus $tHt=H$.

If $g\in G\setminus H$, the even permutation $k=tgt$ moves $\infty$ into $\Omega$. The alternating group $A_n$ is transitive for $n\ge 3$: a suitable 3-cycle carries any prescribed point to another. Choose $u\in G$ with $u(a)=k(\infty )$. Then $v=t{u}^{-1}k$ is even and fixes $\infty$, so $v\in G$ and $k=utv\in GtG$. All conditions of the double-coset criterion for a one-point extension hold, including when $n=3$ and $H$ is trivial.

The constructed extension lies in $A_{n+1}$ and has order $(n+1)|A_n|=(n+1)!/2$. Therefore