The following corollary is on page number 572 in the 14th chapter Galois Theory:
Corollary 10. Let $K/F$ be any finite extension. Then $$|Aut(K/F)| \le [ K : F ] $$ with equality if and only if $F$ is the fixed field of $Aut(K/F)$. Put another way, $K/F$ is Galois if and only if $F$ is the fixed field of $Aut(K/F)$.
The following is definition of $Aut(K/F)$ from the book on page number 558:
Definition. Let $K/F$ be an extension of fields. Let $Aut(K/F)$ be the collection of automorphisms of $K$ which fix $F$.
The definition of fixed field is given on page number 560 as:
Definition. If $H$ is a subgroup of the group of automorphisms of $K$, the subfield of $K$ fixed by all the elements of $H$ is called the fixed field of $H$.
The above corollary belongs to the theorem which is given on page number 570:
Theorem 9. Let $G=\{\sigma_1 =1,\sigma_2,...,\sigma_n \}$ be a subgroup of automorphisms of a field $K$ and let $F$ be the fixed field. Then $$[K:F]=n=|G|.$$
Coming back to the corollary, if by definition of $Aut(K/F)$ the group $Aut(K/F)$ has elements which fix $F$ then why "...if and only if $F$ is the fixed field of $Aut(K/F)$" is included in the statement? (Note: It is proven in the book that $Aut(K/F)$ is a group.)
Also all the elements of $Aut(K/F)$ fix $F$ so by definition of fixed field, $F$ is the fixed field of $Aut(K/F)$. So isn't it that $|Aut(K/F)|=[K:F]$ straight away?