Conjugacy class of $G$ splitting as Conjugacy class of $H\unlhd G$

$\newcommand{\Size}[1]{\lvert #1 \rvert}$It is probably useful to generalize this slightly. You have a group $G$ acting transitively on a set $A$, and a normal subgroup $H$ of $G$. Let $a \in A$.

We have $$ \lvert G \rvert = \lvert a^G \rvert \cdot \lvert G_a \rvert, \qquad \lvert H \rvert = \lvert a^H \rvert \cdot \lvert H_a \rvert. $$ (Here $a^G$ is the orbit of $a$ under $G$, and $G_a$ the stabilizer of $a$ in $G$.)

Also, $H_a = H \cap G_a$, and $$ \frac{H G_a}{H} \cong \frac{G_a}{H \cap G_a}. $$ Thus $$ \frac{\Size{a^G}}{\Size{a^H}} = \frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H_a}}{\Size{G_a}} = \frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H \cap G_a}}{\Size{G_a}} = \frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H}}{\Size{H G_a}} = \Size{G : H G_a}. $$

Note also the following argument. Let $a, b \in A$, with $b = a^g$, for some $g \in G$. Then $$ b^H = a^{g H} = (a^{g H g^{-1}})^g = (a^H)^g, $$ (we have used the fact that $H$ is normal in $G$) which shows that $g$ maps the orbit $a^H$ onto the orbit $b^H$, hence they have the same number of elements.

Define an action of $\;G\;$ on $\;H\;$ by conjugation:

$$g\cdot h\mapsto h^g:=g^{-1}hg\;\;,\;\;g\in G\;,\;\;h\in H$$

What's the order of an orbit of an element here?

$$\left|\mathcal O(h)\right|:=\left|\{h^g\;;\;g\in G\}\right|=[G:G_h]\;,\;;\text{with}\;\;G_h:=\{g\in G\;;\;h^g=h\}=:C_G(h)$$

Of course, we have $\;\mathcal K=\mathcal O(x)\subset H\;$ , which btw means also that $\;x\in H\;$ ...

Now let $\;H\;$ act on itself by conjugation, so that

$$\left|\mathcal O_H(x)\right|:=\left|\{x^h\;;\;h\in H\}\right|=[H:H_x]$$

But

$$y\in H_x\iff x^y=x\iff y\in H\cap C_G(x)$$

and by the (second, third, something) isomorphism theorem:

$$[H:H_x]=\left|H/(H\cap C_G(x))\right|\cong \left|HC_G(x)/C_G(x)\right|=[HC_G(x):C_G(x)]$$

Putting the above together: