Is inversion ever a field automorphism? Is the function sending every element of a field to its multiplicative inverse (and 0 to 0) ever a field automorphism?

Proposition. Inversion is an automorphism of a field $F$ if and and only if $F$ is one of $\mathbb F_2$, $\mathbb F_3$, or $\mathbb F_4$.
Proof:
Let $F$ be a field and suppose that inversion is a field automorphism of $F$. If $x \in F \setminus \{0,1\}$, then $x$ and $1-x$ are both invertible, and hence, by our assumption,
$$
1=1^{-1}=(x+(1-x))^{-1}=x^{-1} + (1-x)^{-1}.
$$
Multiplying out, it follows that $x(1-x)=(1-x) + x=1$, and hence
$$
x^2-x+1=0.
$$
Since a quadratic polynomial over a field has at most two roots, this means that there are at most two possibilities for $x$. Hence $F$ has at most $4$ elements.
If $F=\mathbb F_2$ or $F=\mathbb F_3$, then inversion is the identity map, and hence an automorphism. If $F=\mathbb F_4$, there is only one non-trivial field automorphism $\varphi$, namely the one which satisfies $\varphi(x)=x^2$. We know that multiplicative group $\mathbb F_4^\times$ is cyclic of order $3$. The field automorphism $\varphi$, when restricted to $\mathbb F_4^\times$, induces an automorphism of the multiplicative group $\mathbb F_4^\times$, which must also be non-trivial. However, the only non-trivial automorphism of a cyclic group of order $3$ is inversion. Hence $\varphi$ is the inversion automorphism. (Of course, one can also check this directly for $\mathbb F_4$, if one is more comfortable doing the explicit calculation.)
