Order of Conjugate of Subgroup

Theorem
Let $$G$$ be a group.

Let $$H$$ be a subgroup of $$G$$ such that $$H$$ is finite.

Then $$\left|{H^a}\right| = \left|{H}\right|$$.

Proof
From the definition of Conjugate of a Set we have $$H^a = a H a^{-1}$$.

From Set Equivalence of Regular Representations, $$\left|{a H a^{-1}}\right| = \left|{a H}\right| = \left|{H}\right|$$.