Let $k$ be a field and $K$ be a finite normal extension of $k$. Let $G$ be the group of automorphisms of $K$ over $k$ (since $K/k$ is not necessarily separable, this group may not be called Galois group). Let $f(X)$ be an irreducible polynomial in $k[X]$. I need to show that $G$ acts transitively on irreducible factors of $f$ in $K[X]$, that is, if $g,h$ are irreducible factors of $f(X)$ in $K[X]$, there exists an automorphism $\sigma \in G$ such that $g=h^\sigma$.
Here is my attempt and somehow I'm stuck at the last step. Those trivial cases are not considered (for example, when $g=h$ or when $f$ splits over $K$). Notice that $h^\sigma(X)$ is a factor of $f(X)$ for all $\sigma \in G$. Thus we consider $F(X)=\prod_{\sigma \in G}h^\sigma$. Then $F(X)$ is stable under the action of $G$. The proof would've been finished if I can show that $F=f$.
But I cannot go on because $F$ being stable under $G$ does not imply that $F(X) \in k[X]$, but only $F(X) \in K^G[X]$, where $K^G$ is the fixed field of $K$ under $G$. This is because, as the title indicates, we are working in the context of normal extension, which is not necessarily separable and hence Galois. We only know that $K^G/k$ is a purely inseparable extension and $K/K^G$ is separable. I wonder how can I pass the argument from $K^G$ to $k$, or if my attempt is valid at all.
Any hint or explanation will be appreciated.