Show that there exists an $F$-isomorphism $\sigma : K \to K$ that takes $g$ to $h$ 
Let $K|F$ be a finite normal extension and $f(x) \in F[x]$ be irreducible. Let $g,h \in K[x]$ be two irreducible factors of $f$ in $K[x]$. Show that there exists an $F$-isomorphism $\sigma : K \to K$ that takes $g$ to $h$.

I can prove it by using certain results in splitting fields if we assume that $g$ and $h$ have same degree. By considering some examples, it seems to me that this assumption is correct. But I can't prove it. Is this assumption correct? If yes, how to prove it? If no, how to prove the original statement?
PS: Please refrain from using Galois Theory
 A: Fix an algebraic closure $\Phi$ of $F$ which contains $K$. Let $\alpha\in\Phi$ be a root of $g$ and let $\beta\in\Phi$ be a root of $h$. Since $\alpha, \beta$ have the same minimal polynomial over $F$, there is an $F$-embedding $F(\alpha)\to\Phi$ which sends $\alpha\to\beta$. The extension $\Phi/F(\alpha)$ is algebraic and so we can extend this embedding to a homomorphism $\sigma:\Phi\to\Phi$. Since $K/F$ is a normal extension, the restriction $\sigma|_K$ is an automorphism $K\to K$ over $F$. We'll show this is the automorphism we were looking for.
$\sigma|_K$ can be extended to an automorphism of the polynomial ring $K[x]$ by $\sigma(\sum\limits_{i=0}^n a_ix^i)=\sum\limits_{i=0}^n \sigma(a_i)x^i$. Clearly this automorphism preserves irreducibility over $K$. Also, $\sigma(f)=f$, because $f\in F$, and so its coefficients are fixed by $\sigma$. If so, $\sigma(g)$ must be an irreducible factor of $f$ over $K$. Since $\alpha$ is a root of $g$, it follows that $\sigma(\alpha)=\beta$ is a root of $\sigma(g)$. But the only irreducible factor of $f$ which has $\beta$ as a root is $h$, and so $\sigma(g)=h$, as desired.
Note that in particular it follows that all the irreducible components of $f$ over $K$ have the same degree. Recall that one of the definitions of a normal extension states that if an irreducible $f\in F[x]$ has a root in $K$ then it splits in $K$. So this exercise generalizes this.
