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
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.