Given a transitive action of a group $G$ on a set $X$ and $H\le G$, what's a nec+suff condition for all stabilizers of $X$ in $H$ to be $H$-conjugate?

Question

I have a transitive action of a group $G$ on a finite set $X$.

I also have a subgroup $H\le G$, which may act with many orbits.

I'm looking for a necessary and sufficient condition for all stabilizers: $$\text{Stab}_H(x) \qquad x\in X$$ to be conjugate in $H$.

A sufficient condition is certainly that the centralizer $C_{S_X}(H)$ of $H$ in the symmetric group $S_X$ on $X$ to act transitively on the $H$-orbits, or equivalently for the subgroup $H\cdot C_{S_X}(H)\subset S_X$ to generate a transitive subgroup, but I don't think this condition is necessary.

Answer 1

If $X$ is transitive then it is isomorphic as a $G$-set to the coset space $G/K$, where $K$ is any of the (necessarily conjugate) point-stabilizers in $G$. When $H$ acts on $G/K$, the stabilizer of $gK$ will have elements with $h(gK)=gK$ or equivalently $h\in gKg^{-1}$... so $\mathrm{Stab}_H(gK)=H\cap {}^gK$.

The $\mathrm{Stab}_H(x)$s are all conjugate if and only if each $H\cap{}^gK$ is conjugate to $H\cap K$ in $H$. Thus, given any $g\in G$, here exists an $h\in H$ such that $H\cap{}^gK=H\cap {}^hK$.

Clearly $G=HN_G(K)$ is sufficient for this, but is it also necessary? I don't think so. Let $N$ be the group of all $n\in G$ such that $H\cap{}^nK=H\cap K$ (a kind of "relative normalizer"). Then $G=HN$ is necessary and sufficient. Dunno if there's something better than that to say.