Exercise of conjugacy classes respect a normal subgroup. I'm self-studying some group theory and had troubles with the following exercise.
Assume that $H$ is a normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x \in \mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k = | G : HC_{G} (x) | $.
It has a hint:
Let $A = C_{G} (x)$ and $B= H$ so $A \cap B = C_{H} (x)$. Draw the lattice diagram associated to the Second Isomorphism Theorem and interpret the appropiate indices.
I have seen that as $H$ is normal, you have $N_{G} (H) = G$ so $A \leq N_{G} (B)$ and you can apply the theorem. But didn't made any progress. I checked that if the class $\mathcal{K}$ of $x$ in $G$ is the same as its conjugation class in $H$, then $H C_{G} (x) = G$ and for this easy case it's proven. But I don't think I can generalize my argument.
Any tips?
 A: For readability I'll write $C$ for $C_G(x)$. I don't need the hint but use the orbit-stabilizer theorem.


*

*Show that the orbits of $H$  in $\mathcal{K}$ all have the same size by showing that they have isomorphic stabilizers.

*Show that two elements belong to the same orbit $\iff$ they are $H$-conjugated<.

*Consider the action of $G$ on these orbits (interpreted as"points") and calculate the stabilizer of the orbit corresponding to $H$. This means the subgroup
that "fixes" this orbit in the sense that it permutes elements of this orbits internally.

*Conclude that this group is $CH$.

A: I don't know if answering this way is the correct way, but I couldn't fit it in a comment of Bogaerts Marc's answer.
Let's see if I got it somehow right.
I couldn't do 1.


*If $x,y \in \mathcal{O}$ then $h \cdot x = y$, that is $h x h^{-1} = y$. Also if $h x h^{-1} = y$ then they're in the same orbit as $\mathcal{O_{x}} = \{ a \in G / \: \exists g \in G \: \text{ with } \: g x g^{-1} = a   \}$

*I didn't really understand "the orbit corresponding to "$H$" as "$H$" is normal so if it had orbits it would be itself. Did you meant the orbit corresponding to $\mathcal{O}_{x}$? At least in this way I think I understood a way to arrive to a solution
If it's so, as $\mathcal{K} = \{ gxg^{-1}, g \in G \}$ then we have that this action is transitive in $\mathcal{K}$ so $\mathcal{K}$ is the union of the conjugates of $\mathcal{O}_{x}$. Then
$Orb(O_{x}) = G : Stab(O_{x})$ and if $CH = Stab(\mathcal{O}_{x})$ we're done.
Let's see that $C \subset Stab(\mathcal{O}_{x})$ and $H \subset Stab(O_{x})$
If $g \in C = C_{G} (x)$ then $g x g^{-1} = x$ and so $g \mathcal{O}_{x} g^{-1} = O_{x}$ as they're equivalence classes so they coincide if they have non-empty intersection.
If $g \in H$ then $g \mathcal{O}_{x} g^{-1} = \mathcal{O}_{x}$ as they're orbits on $H$.
As we know that $H C_{G} (x)$ is a group by the second isomorphism theorem, then we have $H C_{G} (x) \subset Stab(\mathcal{O}_{x})$ and we only have to prove that the inclusion holds the other way too.
