Frobenius maps and irreducible functions on finite fields

Question

Let $\mathbb{F}_q$ be a finite field of order $q=p^n$ for some prime $p$ and $n>1$. Suppose both $f(x)=x^2-ax+b$ and $g(x)=x^2-a'x+b'$ are both irreducible.

If, assuming that either $a=a'=0$ or $a,a'\neq 0$, can we conclude $\exists\alpha \in$ Aut$(\mathbb{F}_q)$ such that $\alpha(a)=a'$ and $\alpha(b)=b'$ so that $g(x)=x^2-\alpha(a)x+\alpha(b)$?

Updated Questions

1. If $g(x)=x^2+a'x+b'$ is irreducible, does there exist $f(x)=x^2+ax+b$ irreducible and a non trivial $\alpha$ such that $\alpha(a)=a'$ and $\alpha(b)=b'$?
2. If $g(x)=x^2+a^kx+b^k$ is irreducible, then must $k=p^i$?

Answer 1

No.

There are $q^2$ degree $2$ monic polynomials, and $q(q+1)/2$ products of monic linear factors, which leaves $q(q-1)/2$ monic irreducible polynomials of degree $2$.

Meanwhile, the group of automorphisms of $\Bbb F_q$ has cardinal $n$, so there is no way that the nearly $p^{2n}/2$ different irreducible polynomials can be in a single orbit of size $n$.

Even if you restrict to $a=a'=0$ this leaves you with $(q-1)/2$ polynomials (for odd $q$, there is none if $p=2$ anyway), which is not much closer to $n$.

Answer 2

Answering the updated parts.

1. Yes. For each non-trivial $\alpha$ the polynomial $f(x)=x^2+\alpha^{-1}(a')x+\alpha^{-1}(b')$ is irreducible whenever $g(x)=x^2+a'x+b'$ is. Applying the same automorphism to all the coefficients of an irreducible polynomial is an automorphism of the polynomial ring, and hence maps any irreducible polynomial to another irreducible polynomial.
2. No. If you really meant this to be about $g(x)$ alone, then it is obviously "No", because by using $k$ such that $\gcd(k,p^n-1)=1$ we can use surjectivity of the mapping $x\mapsto x^k$ to get any polynomial we want written in that form. If you really meant that both $f(x)=x^2+ax+b$ and $g(x)=x^2+a^kx+b^k$ should be irreducible then the answer is still "No". A simple example is when $p>2$, $a=0$, and $b$ is a non-square. Then the polynomial $f(x)=x^2-b$ is irreducible. As is $g(x)=x^2-b^k$ for any odd exponent $k$.