Find the orbit and stabilizer of each element of a set of subgroups Let $S_3$ be the symmetric group of all permutations on the set $\{1, 2, 3\}$. Then, let $S$ be the set of all subgroups of $S_3$. Consider $S$ as an $S_3$-set with respect to conjugation and for each element of $S$ find its orbit and its stabilizer.
This is a problem I am working on from an exercise in my notes for algebra so I have the answer to this, however I have been unable to arrive at the correct answer and can't understand where my understanding is falling short.
What I have so far:
We know that $S_3$ is the set $\{e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)\}$ and so I was fairly easily able to deduct that the set of all subgroups $S = \{\{e\}, S_3, \{e, (1 2)\}, \{e, (1 3)\}, \{e, (2 3)\}, \{e, (1 2 3), (1 3 2)\}\}$.
I then named each element of $S$ as $H_1:H_6$, that is $H_1 = \{e\}$, $H_2 = S_3$, $H_3 = \{e, (1 2)\}$, ...
I then used the definition of orbit to write $S_3H_1 = \{\alpha H_1 : \alpha \in S_3\}$,  which is to be repeated for all of the other elements of $S$.
And I used the definition of the stabilizer to write $S_{3_{H_1}} = \{\alpha \in S_3 : \alpha H_1 = H_1\}$, which is also to be repeated for all of the other elements of $S$.
Each time I have tried to work these out, however, I have not achieved the correct answer. I wonder whether the reason why is in the part of the question that tells us to "consider $S$ as an $S_3$-set with respect to conjugation".
Thanks for any help!

Hint:  There are four  different orbits, owing to the fact that the permutations of the same cycle type are all conjugates. And two of them
(and corresponding stabilizers) are obvious,  namely the ones corresponding to the identity subgroup and the one corresponding to  $S_3$.
