This is on Dummit and Foote P134.
If $H\cong\Bbb Z_2$, then since $H$ has unique elements of orders $1$ and $2$, corollary 14 forces ${\rm Aut}(H)=1$.
Corollary 14: If $K$ is any subgroup of group $G$ and $g\in G$, then $K\cong gKg^{-1}$. Conjugate elements and conjugate subgroups have the same order.
I can see since ${\rm Aut}(\Bbb Z_2)$ has only trivial automorphism, and $H\cong\Bbb Z_2$, so ${\rm Aut}(H)=1$.
But I can not see how corollary 14 showed that, since to me, corollary 14 only showed that the conjugation is trivial, and didn't show that ${\rm Aut}(H)$ is trivial.
