This appears in a bigger context, but in summary, I have $K\subset L$ extension of finite fields and $\alpha\in L$ such that $L=K(\alpha)$. Also, $G=Aut_K L$. Now, I need to prove the following:
$$f_K^{\alpha}= \prod_{\sigma\in G} (X-\sigma(\alpha)) $$
But I think that’s pretty easy once you prove $g= \prod_{\sigma\in G} (X-\sigma(\alpha))\in K[X] $. How should I do it?
For simplicity, denote by $f$ the minimal polynomial of $\alpha$ over $K$. Note that your statement is equivalent to that the set of all roots of $f$ is exactly the set $\{\sigma(\alpha):\sigma\in\text{Aut}_K(L)\}$.
First, for any $\sigma\in\text{Aut}_K(L)$, we have $f(\sigma(\alpha))=\sigma(f(\alpha))=0$ because $\sigma$ fixes $K$. Conversely, let $\beta$ be a root of $f$ in an algebraic closure of $K$ containing $L$. Since $\alpha$ and $\beta$ are conjugates, there is an isomorphism $\varphi:L=K(\alpha)\to K(\beta)$ fixing $K$. This shows that $K(\beta)\cong L$. Now, since any extension between finite fields is a normal extension, the fact that $\alpha\in L$ implies that $\beta\in L$, so we have $K(\beta)\le L$. Since $K(\beta)$ and $L$ are both finite-dimensional vector spaces, we then have $K(\beta)=L$. We have now shown that $\varphi$ is in fact an automorphism of $L$ fixing $K$, and $\varphi(\alpha)=\beta$, as desired.