Q:

Orbits under action of a subgroup on the set of conjagtes of a second subgroup

i have the following question:
Let $A\leq B\leq G$ be finite groups. Then $G$ acts naturally via conjugation on the set of conjugates $A^G$. It's trivial, that there is only one orbit under this action. But the action of $G$ on $A^G$ induces an action of $B$ on the set $A^G$. Assume, that there are at least two different orbits $(A^{g_1})^B$ and $(A^{g_2})^B$ under the action of $B$ on $A^G$ for some $A^{g_1},A^{g_2}\in A^G$. Do we know anything about the cardinalities of $|(A^{g_1})^B|$ and $|(A^{g_2})^B|$?
I am especially interested in the case, where $A^{g_1},A^{g_2}\leq B$: Does then hold $|(A^{g_1})^B|=|(A^{g_2})^B|$?
I don't think, that one can show this equality. So maybe someone has an idea for a counterexample? Maybe a semidirect product of two groups, where one of these two groups is itself a semidirect product?
Thanking you in anticipation!

A:

Here is a small example of a group with a group that does not control its own fusion:
Let $G = S_4$, $B=D_8$ be a Sylow 2-subgroup, and $A=Z(B)$. Then $A^G$ has three elements, all contained in $B$, but obviously $A^B$ has only one element $A$. The other two elements of $A^G$ lie in the other orbit.

A (different) general method: Let $G_1$ be a non-regular permutation group and $B_1$ be a point stabilizer (or any non-transitive subgroup with orbits of different sizes). Then $G = A \wr G_1$ and $B =  A \wr B_1$ has the property you want.

