For every field $F$, $\operatorname{id}_{F}$ is an automorphism of $F$.
If $\sigma$ is an automorphism of $F$, then $\sigma(1) = 1$, and $\sigma(1 + 1 + \cdots + 1) = 1 + 1 + \cdots + 1$. Hence, $\sigma$ fixes the prime field of $F$.
Let $F$ be a field of characteristic $p$ and $\mathrm{Fr} : F \to F$ be the Frobenius endomorphism (sending $a \in F$ to $a^{p}$). $\mathrm{Fr}$ is a field homomorphism, so it is injective. It is surjective if and only if every element in $F$ is a $p$-th power, i.e. if and only if $F$ is perfect by Proposition 1902. Thus, $\mathrm{Fr}$ is an automorphism if and only if the field is perfect.