This raised up while trying to answer this post.
Let $G$ be a group and $H\le G$. Let be $X:=\{aH, a\in G\}$ and $Y:=\{Ha, a\in G \}$. The maps $\circ: G\times X\to X$ and $\star:G\times Y\to Y$, defined respectively by $g\circ aH:=gaH$ and $g\star Hb:=Hbg^{-1}$, are both left transitive actions, with pointwise stabilizers $\mathrm{Stab}_\circ(aH)=aHa^{-1}$ and $\mathrm{Stab}_\star(Hb)=b^{-1}Hb$, respectively. Is there any group action-theoretic argument to conclude that, if the two orbits have prime size (i.e. $[G:H]=p$) and intersect in more than one point (i.e. $\tilde aH=H\tilde a$ for some $\tilde a\in G\setminus H$), then they coincide (i.e. $H\lhd G$)?
Yes, with the same argument: If $aH=Ha$, then $a$ normalizes $H$, so the normalizer of $H$ is strictly larger than $H$ itself. But as the index is prime, the only larger subgroup is $G$.