Prove that if $H \leq G$ has exactly $k$ conjugates in $G$ then there exists a homomoprihsm $\alpha: G \rightarrow S_k$ such that $k$ divides $im(\alpha)$
Proof: Let $G$ act on the set of conjugates of $H$ in $G$. This induces a homomorphism $\alpha: G \rightarrow S_k$
If we can find an element of $g$ that maps to an element of order $k$ in $S_k$ we are done.
Now, the action of $G$ on the set of conjugates of $H$ is transitive. Let us denote these conjugates by the set $\{a_1,a_2,....a_k\}$
Then $\exists g_2 \in G$ s.t. $\alpha(g) = (1,2)$
and
$\exists g_3 \in G$ s.t. $\alpha(g) = (1,3)$
$\exists g_4 \in G$ s.t. $\alpha(g) = (1,4)$
....
$\exists g_k \in G$ s.t. $\alpha(g) = (1,k)$
Then $\alpha(g_k....g_3g_2)=(1,k)...(1,3)(1,2)$
which is an element of order $k$ in $im(\alpha)$
Hint:
Use orbit-stabilizer and the first isomorphism theorem.
Recall that if $K\le H \le G$, then
$$[G:K] = [G:H][H:K].$$
Now that you have the problem mostly worked out in the comments, I want to give a full solution here in case that helps.
Let $G$ act on the set of conjugates of $H$ by conjugation. Set $O_H$ to be the orbit of $H$, and $G_H$ to be the stabilizer subgroup of $H$. Note that since this is a transitive action, $|O_H| = k$, and thus
$$|G/G_H| = [G:G_H] = |O_H| = k.$$
Now, we want to understand $|\operatorname{im}(\alpha)| = |G/\ker(\alpha)| = [G:\ker(\alpha)]$. Note that if $g\in\ker(\alpha)$ then $g$ stabilizes all conjugates of $H$ — and in particular $H$ itself — and thus $g\in G_H$. Therefore, we see that $\ker(\alpha)\le G_H$. This implies that
$$|\operatorname{im}(\alpha)| = [G:\ker(\alpha)] = [G:G_H][G_H:\ker(\alpha)] = k\cdot[G_H:\ker(\alpha)].$$
This implies that $k$ divides $|\operatorname{im}(\alpha)|$.