Conjugation by a fixed group element is an automorphism

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.17
Let $G$ be a group and let $G$ act on itself by left conjugation; i.e., $g \cdot x = gxg^{-1}$. For a fixed $g \in G$, prove that conjugation by $g$ is an automorphism of $G$. Deduce that $x$ and $gxg^{-1}$ have the same order for all $x \in G$ and that for any subset $A \subset G$, $|A| = |gAg^{-1}|$. Here $gAg^{-1} = \{ gag^{-1} \ |\ a \in A \}$.

Solution: Fix $g \in G$. The action of $g$ on $G$ corresponds to some element $\sigma_g \in S_G$; thus conjugation by $g$ is an automorphism of $G$.
We saw in a previous section that if $\varphi : G \rightarrow H$ is an isomorphism, then $|x| = |\varphi(x)|$ for all $x \in G$; in this case, we have $|x| = |gxg^{-1}|$ for all $x \in G$.
Moreover, for any set $A \subseteq G$, the restriction $\sigma_g|_A$ is by definition a bijection $A \rightarrow gAg^{-1}$, so that in particular $|A| = |gAg^{-1}|$.