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

 A: 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.
A: Let $E = F(c_1, \ldots, c_k)$ and $f(X) = (X - a_1)\cdots (X - a_n)(X - b_1)\cdots (X - b_m)$ where $$h(X) = (X - a_1)\cdots (X - a_n)  \text{ and } g(X) = (X - b_1)\cdots (X - b_m) (\text{ also assume } m > n).$$ Let $L$ be the splitting field of $f(X)$. Let $\sigma $ be an automorphism such that it takes $a_i \mapsto b_i$ for $i \leq n$ and $b_j \mapsto b_j$ for $n < j < m$. Since $E$ is a normal extension, it is a well know fact that $\sigma(E) = E$. Let $\sigma(h(X)) = g'(X)$. Then it follows that $g'(X)$ divides $g(X)$ and hence $g(X)$ is not irreducible- a contradiction unless $m = n$.
Remark: Existence of $\sigma:$
Over $F$, we have $F(a_i) \simeq F(b_j)$ as both $a_i$ and $b_i$ are roots  of $f(X)$ with $a_i \mapsto b_i$. Therefore proceding similarly we have $F(a_1, \ldots, a_n) \simeq F(b_1, \ldots, b_n)$ with $a_i \mapsto b_i$ for $i \leq n$ and hence $F(a_1, \ldots a_n, b_1, \ldots, b_m, c_1, \ldots, c_k) \simeq F(a_1, \ldots a_n, b_1, \ldots, b_m, c_1, \ldots, c_k)$ where $a_i \mapsto b_i$ for $i \leq n$, $b_j \mapsto b_j$ for $n < j < m$ and $c_t \mapsto c_t$. 
