Let $G$ be a transitive group which contains a regular Abelian group $H$. Determine a necessary condition for all orbital digraphs of $G$ to be self-paired.
The orbitals of $G$ with edge sets $\{g(u,v) : g \in G\}$ and $\{g(v,u) : g \in G\}$ are called paired orbitals. If these two edge sets are equal, the orbital is considered self-paired.
I have thought about this question for a long time now and this is what I have so far: Since $H$ is regular, $Stab_H(x) = 1 \forall x \in H$. I proved earlier that every self-paired orbital digraph is actually a graph, so I know that if I have the arc $(0,s)$ I need the arc $(-s,0)$ for it to be a graph. I just can't seem to connect the dots to finish the proof.
Any help would be appreciative.
$s, s^{-1}$ must be in the same orbit, so we need the permutation (or mapping) $\iota(s) =s^{-1}$ to be a part of $G$. Or, the subgroup generated by $H$ and $\iota$ must be a subgroup of $G$.