If $\operatorname{Aut} G = \{e\}$, show that $G$ is abelian and that every non-identity element of $G$ has
order $2$.
Proposed in "A Friendly Introduction to Group Theory", by Jake Wellens. Exercise 2.6.
We will prove the contrapositive, that is, if either $G$ is not abelian or has an element
of order greater than two, it has a nontrivial automorphism.
Consider now the first of the two cases, and let $g, h \in G$ such that $gh \neq hg$; in this case conjugation by $g$, given by $$ \pi_g (a) = g a g^{-1} $$ constitutes the desired map, since $$ \pi_g (h) = ghg^{-1} \neq hgg^{-1} = h = e(h). $$ Otherwise, multiplying the left- and right-hand sides by $g$ would yield $gh \neq hg$. It is indeed an automorphism, and its inverse is precisely given by $\pi_{g^{-1}}$.
Now, for the second case we may assume $G$ is abelian, as if this was not the case, the previous example would work again. Take now $p$ such that $p^2 \neq e$, i.e., $p \neq p^{-1}$. Notice that due to commutativity, the map $\varphi(g) = g^{-1}$ is an isomorphism, since $$ \varphi(gh) = (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1} = \varphi(g)\varphi(h), $$ and has an inverse which is itself. Plugging $p$ into $\varphi$ finally shows that it differs from the identity automorphism: $$ \varphi(p) = p^{-1} \neq p = e(p), $$ just as we wanted to prove.
Consider now the first of the two cases, and let $g, h \in G$ such that $gh \neq hg$; in this case conjugation by $g$, given by $$ \pi_g (a) = g a g^{-1} $$ constitutes the desired map, since $$ \pi_g (h) = ghg^{-1} \neq hgg^{-1} = h = e(h). $$ Otherwise, multiplying the left- and right-hand sides by $g$ would yield $gh \neq hg$. It is indeed an automorphism, and its inverse is precisely given by $\pi_{g^{-1}}$.
Now, for the second case we may assume $G$ is abelian, as if this was not the case, the previous example would work again. Take now $p$ such that $p^2 \neq e$, i.e., $p \neq p^{-1}$. Notice that due to commutativity, the map $\varphi(g) = g^{-1}$ is an isomorphism, since $$ \varphi(gh) = (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1} = \varphi(g)\varphi(h), $$ and has an inverse which is itself. Plugging $p$ into $\varphi$ finally shows that it differs from the identity automorphism: $$ \varphi(p) = p^{-1} \neq p = e(p), $$ just as we wanted to prove.