This is the exercise 12 in page 111 of the book Algebra from the viewpoint of Galois Theory of Siegfried Bosch:
Let $L/K$ a normal algebraic extension and $f\in K[X]$ monic an irreducible. Let $f=\prod_{k=1}^m f_k$ a prime factorization on $L[X]$ where each $f_k$ is monic. Show that for any two factors $f_i,f_j$ with $i\neq j$ that there is a $K$-automorphism $\sigma :L\to L$ such that $f_i^{\sigma }=f_j$.
I'm not sure if the proof below is correct or there is some mistake, so I post here so anyone can show me if there is a mistake. In the following if $f=\sum_{k=0}^n c_k X^k$ then $f^{\sigma }:=\sum_{k=0}^n \sigma (c_k)X^k$ for some homomorphism of fields $\sigma$.
Let $\overline{L}$ an algebraic closure of $L$ and $\alpha ,\beta \in \overline{L}$ two different roots of $f$. Also let $\rho :K[\alpha ]\to \overline{L}$ the $K$-homomorphism where $\operatorname{img}\rho =K[\beta ]$ and $\rho ':L[\alpha ]\to \overline{L}$ an extension, and finally set $\sigma :=\rho '|_L$. I want to show that if $f_1(\alpha )=f_2(\beta )=0$ then $f^{\sigma }_1=f_2$.
First observe that $f^{\rho' }=f$ as $\rho '$ is a $K$-homomorphism, so $f^{\sigma }=f$. Also notice that $\rho '(\alpha )=\beta$ as it is an extension of $\rho$, so $f^{\rho '}_1(\beta )=f^{\sigma }_1(\beta )=0$ what means that $f_2$ is a factor of $f_1^{\sigma }$ because $\sigma$ restricts to an automorphism on $L$ (because $L$ is normal) so $f=\prod_{k=1}^m f^{\sigma }_k$ and each $f^{\sigma }_k$ is a polynomial in $L[X]$. Moreover, the induced polynomial map $\tilde\sigma :L[X]\to L[X],\, f\mapsto f^{\sigma }$ is an automorphism of polynomial rings as it is invertible because $\sigma$ restricts to an automorphism of $L$, so it preserves irreducibility, from where we conclude that $f_1^{\sigma }=f_2$ as desired.∎