Group action conjugation counting argument Suppose $G$ is a finite group and $H$ is a subgroup. Let $C=C(g)$ be a conjugacy class of $g$ in $G$ and $D$ a conjugacy class in $H$  contained in $C(g)$ (may or may not contain $g$). Show that:
$$\frac{|\{x\in G : xgx^{-1} \in D\}|}{|G|}= \frac{|D|}{|C|}.$$
My attempt:
Note that $|\{ x \in G: xgx^{-1}=g \}|=\frac{|G|}{|C|}$. Thus it suffices to show that
$$|\{ x\in G : xgx^{-1}\in D\}|={\rm Stab}(g)|D|$$
However, I have no idea how to proceed
 A: I will try to phrase things in a way that is most easy to generalize.
Suppose $D$ is the $H$-orbit of $ugu^{-1}$ for some $u\in G$. Writing ${}^xg:=xgx^{-1}$, define
$$ X:=\{x\in G\mid {}^xg\in D\} $$
The condition $xgx^{-1}\in D$ is equivalent to $xgx^{-1}=hugu^{-1}h^{-1}$ for some $h\in H$, which in turn is equivalent to $x\equiv hu$ mod $\mathrm{Stab}_G(g)$ for some $h\in H$ (meaning $x$ and $hu$ represent the same coset of $\mathrm{Stab}_G(g)$, since they act the same on $g$), which in turn is equivalent to $x\in Hu\mathrm{Stab}_G(g)$. This double coset has the same size as $Hu\mathrm{Stab}_G(g)u^{-1}$ which is the same set as $H\mathrm{Stab}_G({}^ug)$, so
$$ |X|=|H\mathrm{Stab}_G({}^ug)| = \frac{\color{DarkOrange}{|H|}\color{Blue}{|\mathrm{Stab}_G({}^ug)|}}{\color{DarkOrange}{|H\cap\mathrm{Stab}_G({}^ug)|}} $$
I will use $K$ for the conjugacy class of $g$ in $G$ instead of $C$ to avoid confusion with centralizers (and I am also referring to stabilizers when, in this case, the stabilizers are centralizers).
By the orbit-stabilizer theorem, we can replace the orange and blue parts as so:
$$ |X|=\color{DarkOrange}{|D|}\cdot\frac{\color{Blue}{|G|}}{\color{Blue}{|K|}} $$
and the result follows.
A: Let my know if I get the question wrong.
As you said $D$ is a conjugacy class (orbit of conjugate action) which is contained in $C(g)$. Note that by this we know that $x \in D$, then $x$ is a conjugate of $g$; But $D$ is an orbit, by definition it means that it must contain all conjugates of all of its elements, particularly $x$, so by symmetry of conjugation, $D$ actually must contain $g$. And again therefore since it contains $g$, it must contain $C(g)$. So $D = C(g)$.
This equality could have been achieved in some other way too: Orbits of an action (here conjugacy classes) yield a partition on the set, so if an orbit is contained in another, it must be equal to it.
And therefore the equation is stating $1=1$.
It's also suspicious that some $H$ is defined, because from the description you've provided, there is no need for $H$ to prove the statement.
