Consider an absolutely irreducible polynomial $f(x)$ with coefficients in $\mathbb{F}_{q^r}$. Consider also $Gal(\mathbb{F}_{q^r}, \mathbb{F}_q)$. Is there any possibility that $\sigma(f(x))$ remains an absolutely irreducible polynomial where $\sigma \in Gal(\mathbb{F}_{q^r}, \mathbb{F}_q)$ different from identity?
To be more clear, my question is that when we take a function $p(x) \in \mathbb{F}_q [x]$ and suppose that $p(x)= \prod_{\sigma \in Gal(\mathbb(F)_{q^r}, \mathbb{F}_q)} \sigma(h(x))$ where $h(x)$ is a polynomial in $\mathbb{F}_{q^r}[x]$ which is absolutely irreducible. I am wondering that can $\sigma(h(x))$ be absolutely irreducible under some condition on $\sigma$? If so, what are the conditions on $\sigma$?
Let $f \in \mathbb{F}_{q^r}[x]$ be absolutely irreducible.
Given a polynomial $g$ and an automorphism $\sigma$ of a field containing the parameters of $g$ (eg a global automorphism), allow me to write $g^\sigma : x \mapsto \sigma(g(x))$. In other words $g^\sigma(x) := \sigma(g(x))$.
For any global automorphism $\sigma \in Gal(\operatorname{acl}(\mathbb{F}_{q}), \mathbb{F}_q)$, the polynomial $f^\sigma$ is also absolutely irreducible.
To see this, notice that if $f^\sigma$ factors, say $f^\sigma = P \cdot Q$ where $P, Q$ have coefficients in the algebraic closure of $\mathbb{F}_q$, then $f = (f^\sigma)^{\sigma^{-1}} = (P \cdot Q)^{\sigma^{-1}} = P^{\sigma^{-1}} \cdot Q^{\sigma^{-1}}$. Moreover, $\deg(P) = \deg(P^{\sigma^{-1}})$ and $\deg(Q) = \deg(Q^{\sigma^{-1}})$.