Is F a field automorphism?

Question

Let $\mathbb{F}_q$ be a finite field of order $q=p^n$ for some prime $p$ and let $f:\mathbb{F}_q\to \mathbb{F}_q$ be a bijection. If

1. $f(0)=0$ (additive identity),
2. $f(1)=1$ (multiplicative identity),
3. $f(a)f(b)=f(ab)$ for all $a,b\in \mathbb{F}_q$,

then is $f$ a field automorphism ($i.e.$ does it follow that $f(a+b)=f(a)+f(b)$)?

A source with a proof or counterexample would be very helpful.

Updated Question:

1. Is every such bijection above that is not an automorphism of the form $f(x)=x^k$ for some k?
2. Is there another property that $f$ could satisfy to force it to be an automorphism?

Answer 1

The map $a\mapsto a^{-1}$ (for $a\ne0$) and $0\mapsto0$ satisfies the requirement, but is not generally a field automorphism: $$ a^{-1}+b^{-1}=\frac{a+b}{ab} $$ so the equality $$ (a+b)^{-1}=\frac{a+b}{ab} $$ should hold as soon as $a+b\ne0$; this means $$ (a+b)^2=ab $$ that is $$ a^2+ab+b^2=0 $$ in particular $3=0$.

So if $p\ne2$ and $p\ne3$ we're done. Try your hand for the other cases.

Answer 2

There are $n$ field automorphisms of $\mathbb{F}_q$, since it is a Galois extension of degree $n$ of $\mathbb{F}_p$.

On the other hand, your $f$ corresponds to any group automorphism of $\mathbb{F}_q^* \simeq \mathbb{Z}/(q-1)\mathbb{Z}$ (extended by $0\mapsto 0$). Now there are $\phi(q-1)$ such automorphisms (with $\phi$ the Euler function).

So your question amounts to whether $\phi(p^n-1)>n$. I'm not sure if it always holds, but at least it does for big enough $p$ and $n$.