Source: Abstract Algebra, 3rd edition, Dummit and Foote.
In section 4.4, the book goes: "For example, if $H\cong Z_2$, then since $H$ has unique elements of order $1$ and $2$, Corollary 14 forces $\text{Aut}(H)=1$."
Here is Corollary 14: "If $K$ is any subgroup of the group $G$ and $g\in G$, then $K\cong gKg^{-1}$. Conjugate elements and conjugate subgroups have the same order."
My Question: It is clear as day that $\text{Aut}(H)=1$ if $H\cong Z_2$; but I couldn't for the life of me follow how exactly "Corollary 14 FORCES $\text{Aut}(H)=1$". Any hint would be greatly appreciated.
I'll assume $Aut$ means automorphisms induced by conjugation.
Corollary 14 says that an element of order 2 must be mapped to an element of order 2. But we just said that there is only one such element, so any automorphism must map it to itself. And the same holds for the unique element of order 1. So the only automorphism is the identity.
I'll say that this seems like massive overkill, and it's not at all clear to me whether "$Aut(H) = 1$" is supposed to mean "there is only one automorphism", or "$1$" is supposed to represent the identity map. I'm used to $Aut()$ being a set, not a count, so I would have written this as $Aut(H) = \{id\}$.