Note: I'm using Emil Artin's free text on Galois Theory to understand the Fundamental Theorem of Galois Theory.
Hypothesis: Let $F \subset E$ s.t. $E$ is the splitting field for separable $p(x) \in F[x]$. Let $F \subset B \subset E$.
Question: At the beginning of the very last paragraph on page 48, Artin states that
$B$ is a normal extension of $F$ if and only if each isomorphism of $B$ into $E$ is an automorphism of $B$. This follows from the fact that each of the above conditions are equivalent to the assertion that there are the same number of isomorphisms and automorphisms.
This isn't obvious to me at all. How do I make sense of this?
What I DO understand that might be relevant:
I do understand that if $B$ is normal over $F$, then
$$ \underbrace{|G(B:F)| = (B/F)}_{\text{since $B$ normal over $F$}} = [G(E:F) : G_B] $$
I also understand that the elements of $\{\sigma G_B : \sigma \in G(E : F)\}$ can be viewed as equivalence classes of distinct $B$ isomorphisms (when the domain's of the elements of these equivalence classes are restricted to $B$).
I think Artin's text, while legendary, is not suitable for today's audience. There are not many examples, and the notation he used was out of dated. Maybe you should check a relatively common one like Dummit and Foote. I wrote this from the perspective that I learned Galois theory from Artin's text myself for the first time.
For your question, you can think a normal extension as the splitting field of $f(x)$ over $F$ in $E$. Each isomorphism that maps $B$ to $B_{i}$ in $E$ merely permuted the roots of $f(x)$. Therefore all $B_{i}$ must be the same. This only holds for separable extensions, and for non-separable ones you need a family of $f_{i}$s. However the proof should carry through with small modifications.