From what I have read, the fundamental theorem of Galois theory states that there is a bijection between subfields of a splitting field of a polynomial and subgroups of the Galois group.
One question I have is that why do the automorphisms in the Galois group have to fix the polynomial of the splitting field?
In the case where $B = A(\alpha)$ where $\alpha$ is the root to the minimal polynomial f(x), and the Galois group is $Aut(B/A)$, Is it because transforming the polynomial means A isn't fixed?
Another way to ask this is that do all possible automorphisms of $B$ that fixes $A$ also fixes $f(x)$? If not, then why are those automorphisms not in the galois group?