The only fixed-point free automorphism of order $2$ is $\phi(a)=a^{-1}$(in a finite group)

abstract-algebra proof-verification

The proof looks good, just for completion I will add a proof that $f: a \rightarrow a^{-1}$ is an isomorfism if and only if $G$ is abelian.

If the group is abelian then $f(ab) = (ab)^{-1} = (ba)^{-1} = a^{-1}b^{-1} = f(a)f(b)$.

If the group is not abelian take $x$ and $y$ that do not commute and let $a=x^{-1},b=y^{-1}$ and notice $f(ab) = (x^{-1}y^{-1})^{-1} = yx \neq xy = f(a)f(b) $

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