Suppose that $G = R \rtimes G_{\alpha}$ is a finite permutation group on $\Omega$ where $R$ is the non-abelian subgroup of order $27$ and exponent $3$ (see here). Suppose that $G_{\alpha} \cong D_n$, where $D_n$ denotes the dihedral group of order $2n$ (considering $C_2\times C_2$ as dihedral). Now look at the orbits of $G_{\alpha}$, and suppose we have the orbit $\{\alpha\}$ and three other orbits where $G_{\alpha}$ acts non-regular, one of size $2$ and two of size $n$, the other orbits being regular for $G_{\alpha}$.
Now as $|\Omega| = |R| = 27$ this gives that $$ 27 = 1 + 2 + n + n + 2nk $$ for some $k$ by an orbit decomposition. So we have $n \in \{2,3,4,6,12\}$. Now I want to rule out the case $n = 12$. But I have no idea.
So why is $n = 12$ not possible?
This must have something to do with the orbit decomposition above. Some further facts: We have $Z(R) = \Phi(R)$ has order $3$ and $Z(R)$ is normal in $G$ and partitions $\Omega$ into orbits of size $3$ that are blocks under $G$.
Remark: The action of $G_{\alpha}$ on $\Omega$ is equivalent to the action of $G_{\alpha}$ on $R$. So we could as well dispose the set $\Omega$ and just consider a semidirect product $R \rtimes H$ where $H$ acts on $R$.