Relationship between $|N_G(K)|$ and $|N_H(K)|$ for $K\le H\le G$ Suppose $K\le H\le G$ are finite groups.  I'd like to know when the following equation holds:
$$|N_G(K)|=|N_H(K)|\cdot [G:H]$$
A sufficient condition is the normality of $K$ in $G$.  In general, I'm interested in relationships between the orders of the two normalizers $N_G(K)$ and $N_H(K)$.  I feel like I've seen formulas for this before, but I can't seem to find/recall them at the moment.
 A: A slightly more general setting, as hinted to by Giulio Bresciani, is the following.
You have a group $G$ acting transitively on a set $A$, and a subgroup $H$ of $G$. You want to know when is it that for some $a \in A$ you have
$$
\lvert G_{a} \rvert = \lvert H_{a} \rvert \cdot \lvert G : H \rvert,
$$
where $G_{a}$ is the stabilizer of $a$ in $G$, and similarly for $H_{a}$.
Now $\lvert G \rvert = \lvert a^G \rvert \cdot \lvert G_{a} \rvert$ and $\lvert H \rvert = \lvert a^H \rvert \cdot \lvert H_{a} \rvert$. (Here $a^G$ and $a^H$ are the orbits, and $a^G = A$, as $G$ acts transitively.) Moreover $a^H \subseteq a^G$, so
$$
\frac{\lvert H \rvert}{\lvert H_{a} \rvert} = \lvert a^H \rvert \le \lvert a^G \rvert = \frac{\lvert G \rvert}{\lvert G_{a} \rvert},\tag{ineq}
$$
which is equivalent to
$$
\lvert G_{a} \rvert \le \lvert H_{a} \rvert \cdot \lvert G : H \rvert,\tag{main}
$$
and (ineq) yields that equality holds in (main) if and only if $a^H = a^ G = A$, that is, $H$ acts transitively on $A$.
Of course you then apply it to the action of $G$ by conjugacy on the conjugacy class $A$ of $K$.
A: Call $M=N_G(K)$, the equation is
$$|M|=|M\cap H|\cdot[G:H]$$
or
$$[H:M\cap H]\cdot|M|=[H:M\cap H]\cdot|M\cap H|\cdot[G:H]=|H|\cdot[G:H]=|G|=|M|\cdot[G:M]$$
that is
$$[H:M\cap H]=[G:M]$$
Using right cosets, we have a canonical function $j:H/M\cap H\to G/M$ sending $M\cap Ha$ to $Ma$, you can easily check that $j$ is injective. Hence, our equation is equivalent to the surgectivity of $j$, that is: for all $g\in G$ there exist $h\in H$ such that $Mg=Mh$, or $gh^{-1}\in M$. In the end, it's equivalent to $G=HM$, and this you can interpret as $H$ acting transitively on the conjugacy classes of $K\subset G$.
