Partition of a conjugacy class to conjugacy classes of a normal subgroup

Question

Let $G$ be a group, $H$ be a normal subgroup of $G$, and $O$ a conjugacy class of $G$ contained in $H$.

1. Consider $O = \cup_{i = 1}^{n}O_i$ the partition of O into conjugacy classes of $H$. Show that the size of each $O_i$ is the same
2. Prove that the number $n$ of conjugacy classes which $O$ breaks up to in $H$ divides $[G : H]$

I considered the action of $H$ on $O$ by conjugation and the stabilizer of elements of $O$, but it wasn't clear to me what to do next.

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I have a partial solution to the first problem inspired by this answer.

Consider 1 conjugacy class/orbit in $H$: wlog $O_1$ with representative $h_1$. Consider another conjugacy class/orbit $O_2$ with representative $h_2$, and note that $h_2 = g^{-1}hg$ for some $g$ in $G$.

$$O_2 = \{ hg^{-1}h_1gh^{-1} | h \in H \}$$

Now replace $h$ with $g^{-1}hg$ as $H$ is normal in $G$ and we are ranging over all of $H$.

$$O_2 = \{ (g^{-1}hg)g^{-1}h_1g(g^{-1}h^{-1}g) | h \in H \}$$ $$= \{ g^{-1}hh_1h^{-1}g | h \in H \}$$ $$=g^{-1}O_1g$$

So conjugation by $g$ sends $O_1$ to $O_2$. It's not clear to me that $=g^{-1}O_1g$ doesn't get smaller than $O_1$ though (although I think I can work it out with time).

Answer 1

Part 1 is edited in the question and as Arturo commented, conjugation is a bijection on $G$ so $|O_1| = |g^{-1}O_1g|$.

For part 2, let $o_1$ be an element of $O$ (the orbit in $G$), and $O_1$ be the orbit in H of $o_1$. Let $G_{o_1}$ and $H_{o_1}$ be the stabilizers of $o_1$ in $G$ and $H$ respectively.

$$n = \frac{|O|}{|O_1|} = \frac{\frac{|G|}{|G_{o_1}|}}{\frac{|H|}{|H_{o_1}|}}$$ $$= \frac{|G|}{|H|}\frac{|H_{o_1}|}{|G_{o_1}|}$$ $$= [G : H] \frac{|H_{o_1}|}{|G_{o_1}|}$$

So $n$ is divisible by $[G : H]$.