$\newcommand{\Aut}{\mathrm{Aut}} $
Few Definitions:
Let $E$ be a field and $\Aut(E)$ denote the group of automorphisms of $E$.
Now, Let $S\subset \Aut(E)$
Define:$$\mathscr{F}(S):=\{a\in E\enspace|\sigma(a)=a \enspace\forall \enspace \sigma\in S \}$$
We call $\mathscr{F}(S)$ a fixed field of $S$.
Moreover, It is easy to see that $\mathscr{F}(S)$ is a subfield of $E$.
Question:
If $S\subset \Aut(E)$, then $S\subset \Aut(E/\mathscr{F}(S))$
I know that
$\Aut(E/\mathscr{F}(S))=\{\sigma\in \Aut(E)|\sigma(x)=x\enspace \forall x\in \mathscr{F}(S)\}$
I have to show that $$\forall\tau\in S\implies \tau\in \Aut(E/\mathscr{F}(S))$$
I am unable to realize it. Please help.