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!

Maybe the problem is how you have defined the stabilizer and the orbit. If you have a group $G$ and $a,b\in G$ a conjugation is something of the form $a^{-1}ba$. When you write the stabilizer you write $\alpha H_1= H_1$, but I think it should be $\alpha^{-1} H_1 \alpha= H_1$.
