Let $H$ and $K$ be subgroups of a group $G$. The transporter of $H$ into $K$ is the set of all $g\in G$ that conjugate $H$ into $K$:
$$\operatorname{Tran}_G(H,K)=\{g\in G\mid gHg^{-1}\le K\}$$
Suppose now that $H\le K$ so that $\operatorname{Tran}_G(H,K)$ is a $N_G(K)\times N_G(H)$-set, i.e., for any $(n,m)\in N_G(K)\times N_G(H)$ and $g\in \operatorname{Tran}_G(H,K)$, the element $ngm^{-1}$ is also in $\operatorname{Tran}_G(H,K)$.
When does $\operatorname{Tran}_G(H,K)$ consist of one orbit under this action (necessary and/or sufficient conditions)? Are there examples of $H\le K\le G$ where $\operatorname{Tran}_G(H,K)$ consists of multiple orbits?
Note that in the trivial cases of $H=e$ or $K=G$ the action is transitive. Thanks for any thoughts or references. I've not seen this in the literature before.
$H \unlhd G$ and $K \unlhd G$ are also sufficient conditions for there to be a single orbit. In both cases, the transporter is the whole of $G$. It might be hard to find necessary and sufficient conditions for there being a single orbit.
Here is an example in which there are two orbits. Let $G = S_4$, $H = \langle (1,3)(2,4) \rangle$, and $K = \langle (1,2,3,4),(1,4)(2,3) \rangle$. So $|H| = 2$ and $|K|=8$ and $N_G(H)=N_G(K)=K$. But the conjugates of $H$ in $G$ all lie in $K$, so ${\rm Tran}_G(H,K) = G$. However the orbit of the identity element under the action is just $K$. In fact the other $16$ elements of $G$ comprise a single orbt, so there are two orbits altogether.
I expect there are examples with more than $2$ orbits.