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.