$1$.) I count 16 as well:
$D_{12}$ itself and the identity (trivial) subgroup.
$\langle r\rangle,\langle r^2\rangle, \langle r^3\rangle$
Two isomorphs of $S_3$: $\langle r^2,s\rangle$ and $\langle r^2,rs\rangle$.
Three isomorphs of $V$: $\langle r^3,s\rangle$, $\langle r^3,rs\rangle$ and $\langle r^3,r^2s\rangle$.
Six subgroups generated by a single reflection $\langle r^ks\rangle$ for $k = 0,1,2,3,4,5$.
$2$.) Yes, you want to find the normalizers in $D_{12}$. This is the entire group (as these are all normal) for the first seven subgroups listed, which greatly simplifies things. The isomorphs of $V$ are non-normal but of prime index, so they must be their own normalizers. It is clear that the normalizers of the reflection-generated subgroups of order 2 must be an isomorph of $V$ (the isomorphs of $V$ are abelian so they normalize any subgroup), as $S_3$ has no normal subgroups of order 2 (we are using the fact that the normalizer of a subgroup $H$ in a group $G$ is the largest subgroup of $G$ containing $H,$ in which $H$ is normal).
$3$.) Finding the orbits by brute force is not so bad: the seven normal subgroups will all have a single element of just themselves in their orbits, because they are stabilized by all of $D_{12}$. From Sylow theory (or by inspection), we see the isomorphs of $V$ are all conjugate, so there is another orbit (remember conjugation preserves order of subgroups), and we are left with finding the orbits of the reflection-generated subgroups.
By the orbit-stabilizer theorem (or by examining their generators' (ordinary) conjugacy classes) we have that these must occur in two orbits of three: $\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\}$ and $\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}$
Thus: $G/X = \{\{\{1\}\},\{\langle r\rangle\},\{\langle r^2\rangle\}, \{\langle r^3\rangle\},\{D_{12}\},\{\langle r^2,s\rangle\},\{\langle r^2,rs\rangle\},\{\langle r^3,s\rangle, \langle r^3,rs\rangle,\langle r^3,r^2s\rangle\},\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\},\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}\}.$
(This is a set with 10 elements).
Best Answer
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).