In a normal extension of a field, is there an automorphism that maps irreducible factors of a certain irreducible polynomial?

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Let $F$ be a field, $f(x)$ be an irreducible polynomial in $F[x]$ and $E/F$ be a normal extension. Show that if $g(x)$, $h(x)$ are irreducible factors of $f(x)$ in $E[x]$ then there exists an automorphism $\sigma$ of $E$ over $F$ such that $\sigma(g)=h$. Does this result hold if we do not assume normal extension?

What I've tried so far:

Let $\overline{F}$ be the algebraic closure of $F$. Then, by definition, $f(x)\in F[x]$ splits completely over $\overline{F}$. So

$$f(x)=(x-\alpha_1)\cdots(x-\alpha_n)(x-\beta_1)\cdots(x-\beta_m) (x-\gamma_1)\cdots(x-\gamma_k)$$

Since $g(x)$ and $h(x)$ are irreducible factors of $f(x)\in E[x]$ then we can write, without loss of generality, $n\leq m$ and

$$g(x)=(x-\alpha_1)\cdots(x-\alpha_n)\qquad h(x)=(x-\beta_1)\cdots(x-\beta_m)$$

I want to define a map $\sigma:E\rightarrow E$ which maps $\alpha_i$ to $\beta_i$ (with this I can conclude $\sigma(g)=h$, right?). But the problems are:

1) I don't know $n=m$.

2) I don't know $\alpha_i,\beta_i\in E$.

3) Even if $\alpha_i,\beta_i\in E$, I'd only have a map on a subset of $E$. I don't know if I can extend this map to the whole $E$.

EDIT: The question above can be found on Serge Lang's Algebra, Revised Third Edition, Volme 1, Chapter V, exercise 26, and is stated as:

Let $k$ be a field, $f(X)$ an irreducible polynomial in $k[X]$, and let $K$ be a finite normal extension of $k$. If $g$, $h$ are monic irreducible factors of $f(X)$ in $K[X]$, show that there exists an automorphism $\sigma$ of $K$ over $k$ such that $\sigma = h^\sigma$. Give an example when this conclusion is not valid if $K$ is not normal over $k$.

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What you propose is already close to a full solution: There's an automorphism $\sigma\in Gal(K/F)$, $K$ a splitting field of $f$ over $E$ that sends $\sigma(\alpha_1)=\beta_1$; this works because both elements have $f$ as their minimal polynomial over $F$. Since $E$ is normal, it is invariant under $\sigma$, and thus we can restrict. As you suspected, it then follows that $\sigma(g)=h$ because these are two irreducible polynomials that share a root in an extension field, so they are both equal to the minimal polynomial of that root.