find the orbits of the operation of conjugation of $S_3$ on the group of subgroups of $S_3$ $G = S_3$ I need to examine the operation of conjugation of $G$ on the group of the subgroups of $G$ (meaning $gH = gHg^{-1}$)
here I already got confused from the definition.
Now I need to : 
a) find the group of orbits $O$ of this operation.
b) for each orbit $o \in O$ choose a representative $H \in o$ and calculate $\operatorname{Stab}_G(H)$.
c) check the Orbit-stabilizer theorem on this operation.
I'm really confused from the definitions here. 
What I tried -  If I understand the question correctly - I have H' which is a group that contains 6 subgroups of $S_3$ then i calculate the orbits for each of these groups. I get one orbit for each group. So I guess here the representative (asked in (b) ) is this one orbit that I found. for each of the 6 orbits I get that 
$$\operatorname{Stab}_G(O_i) = S_3$$
so in (c) i get that 
$|H'| = \sum_1^6 [G:\operatorname{Stab}_G(x_i)] =  1+1+1+1+1+1 = 6$ which is correct but I'm sure i did not understand the question correctly . any help will be very appreciated!
 A: First of all I guess what you write as conjunction you mean conjugation (at least this is the name I know). Secondly your $H'$ is the set of subgroups of $S_3$. ($H'$ is not a group!!). So $S_3$ acts on $H'$ by conjugation. As you mentioned correctly $S_3$ contains 6 subgroups, where of course $S_3$, the trivial subgroup $\{1\}$ and the alternating group $A_3$ are three of them. The other three subgroups are easily found (and I guess you have written them down explicitly).
Now, you have to find the orbits of this action. So the "straightforward" check would be to take all six elements of $S_3$ and apply them to all of your six subgroups and see what happens. However you can shorten up a little bit, since you don't need any computation for the three subgroups you "know by name". Indeed, if you conjugate $S_3$ (as an element of $H'$) by any element of $S_3$ (the group which acts) you'll always obtain $S_3$. The same is true for $\{1\}$. Furthermore $A_3$ has index 2 in $S_3$, thus it is a normal subgroup. And, by definition, normal subgroups are exactly those subgroups which are fixed by conjugation. In other words conjugating $A_3$ (as an element of $H'$) with any element of $S_3$ (the group which acts) you 'll always get $A_3$ again. Thus, the orbit of any of this three subgroups (elements of $H'$) is just the subgroup itself.
However trying to do this for your remaining three elements of $H'$ (subgroups of $S_3$) this is not true.
For example you have the two subgroups $A=\{ 1, (13)\}$ and $B=\{ 1, (12)\}$. Is there really no element $g \in S_3$ such that $g A g^{-1} = B$? In total, taking this (easy) consideration to an end you should get that there are $4$ orbits (3 of them contain one element [see 2nd paragraph] and one contains three elements).
A: As you suspect, your answer is not correct.  The first thing you need to list all the subgroups of $S_3$.  Now for each subgroup $H \leq S_3$ and for each $g \in S_3$, you need to compute $gHg^{-1}$.  These conjugate subgroups are the elements of the orbit of $H$.
For example, take $H = \langle (1 \ 2) \rangle \leq S_3$. Now we need to loop over all the $g \in S_3$ and compute $gHg^{-1}$.  If we take $g = 1$ or $g = (1\ 2)$, we just get $gHg^{-1} = H$.  Now consider $g = (1 \ 3)$.  Since
$$
(1\ 3) (1\ 2) (1\ 3)^{-1} = (1\ 3) (1\ 2) (1\ 3) = (2\ 3)
$$
we see that $gHg^{-1} = \langle (2\ 3) \rangle$.  Thus $H = \langle (1\ 2) \rangle$ and $\langle (2\ 3) \rangle$ are both in $O_H$, the orbit of $H$.  What other subgroups can you get by conjugating $H$ by other $g \in S_3$?  What happens for the other subgroups?  Try $H = \langle (1\ 2\ 3) \rangle$ and see how this differs from $\langle (1\ 2) \rangle$.
