Number of Distinct Conjugate Subsets is Index of Normalizer

Theorem
Let $G$ be a group.

Let $S$ be a subset of $G$.

Let $N_G \left({S}\right)$ be the normalizer of $S$ in $G$.

Let $\left[{G : N_G \left({S}\right)}\right]$ be the index of $N_G \left({S}\right)$ in $G$.

The number of distinct subsets of a $G$ which are conjugates of $S \subseteq G$ is $\left[{G : N_G \left({S}\right)}\right]$.

Proof
$S^a = S^b \iff S^{a b^{-1}} = S$ (reference to be determined).

That is, $S^a = S^b \iff a b^{-1} \in N_G \left({S}\right)$, which is equivalent to $a^{-1} \equiv b^{-1} \left({\bmod\, N_G \left({S}\right)}\right)$.

Thus we have a bijection between the class $\mathcal{C} \left({S}\right)$ of subsets of $G$ conjugate to $S$ and the left coset space $G / N_G \left({S}\right)$ given by $S^a \to a^{-1} N_G \left({S}\right)$.

Since $G / N_G \left({S}\right)$ has $\left[{G : N_G \left({S}\right)}\right]$ elements, the result follows.