Let a polynomial $f(x)\in{K[x]}$ splits in $F$ as $f(x)=(x-a_{1})^{m_{1}}...(x-a_{n})^{m_{n}}$ where $a_{i}\in{F}$ are distinct. and $E:=K(b_{0},...,b_{n})$ where $b_{0},...,b_{n}$ are the coefficient of the polynomial $g(x)= (x-a_{1})...(x-a_{n})$. Then show that:
(a) $F$ is Galois over $E$, and
(b) $Aut_{E}F=Aut_{K}F$.
I fixed part (a). For part (b), I know $Aut_{E}F\subset{Aut_{K}F}$, but the other direction is killing me. I asked my professor and he told me it is immediate from (a) but I am still having trouble how is it immediate?