Subgroups, Stabilizers, and Orbits of the Dihedral group D12. Faster way? 

  
*Let $G = D_6$ be the dihedral group of the rigid motions of a regular hexagon and $X = \{H \subseteq G \mid H \leq G\}$. Consider the action of $G$ on $X$ by conjugation, that is, for $g \in G$, $H \in X$:
  $$
  g \cdot H = g H g^{-1}.
$$
  a) Find the stabilizer $G_H$ of $H$, for all $H \in X$.
  
  b) Find the orbit $G(H)$ of $H$, for all $H \in X$.
  
  c) Find the set of all orbits $X/G$.
  

I'm working on this problem (note $D_6 = D_{12}$ in the more common notation). So far I've:


*

*Found all the subgroups of $D_{12}$ (all 16, I hope).

*Found the normalizer of each subgroup to get the stabilizers for each subgroup. (Is this correct since the action is conjugation?) 

*Found the orbits via brute force.

*Found all the cosets of each $H$ and started constructing the orbits.


However, all this is taking an ENORMOUSLY long time. So much so that I feel I'm missing a much more clever solution.
Is there one?
 A: $1$.) 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).
