This is corollary 10, page 585 in Dummit&Foote. It is followed immediately after proving that
Theorem 9: If $G = \{1,\sigma_2,\dots,\sigma_n\} \leq Aut(K)$ with fixed field $F$ then
$$[K:F] = n = |G|$$
Corollary 10: Let $K/F$ be any finite extension. Then $$|Aut(K/F)| \leq [K:F]$$ with equality $\iff F$ is fixed field of $Aut(K/F).$
Here is the proof for reference (it's not important)
Let $F_1$ be fixed field of $Aut(K/F)$ so that $F \subset F_1 \subset K.$ Then $[K:F_1] = | Aut(K/F)|.$ Hence $[K:F] = |Aut(K/F)|[F_1:F] \geq |Aut(K/F)|.$
I have some questions about the notation mainly and what we are proving
Q1: I thought the notation $Aut(K/F)$ means exactly automorphisms with fixed field $F$ by definition. This is the very first thing introduced in the chapter.
Q2: So shouldn't the statement then be reworded to
$$|Aut(K/F_1) \leq [K: F]$$ with equality $\iff F = F_1.$
Q3: This refinement seems to be consistent with corollary 11 that follows which briefly states if $G$ is a finite subgroup of $|Aut(K/F_*)$ with fixed field $F_*$, then the "missing" automorphisms fixing $F_*$ are still in $G$, in other words $G = Aut(K/F_*)$ is the whole group. Do I have this right?
By $\operatorname{Aut}(K/F)$ we mean the group of all automorphisms $\sigma$ of $K$ that fix $F$ (pointwise), i.e. $\sigma(x) = x$ for all $x \in F$. This is not the same as saying that $F$ is the fixed field of every automorphism in $\operatorname{Aut}(K/F)$ because it is okay if these automorphisms fix additional elements of $K$ that are not in $F$. For example, the identity morphism is always an element of $\operatorname{Aut}(K/F)$ and it fixes every element of $K$ regardless of whether it is in $F$ or not. So the fixed field of an element $\sigma \in \operatorname{Aut}(K/F)$ contains $F$ but it might be bigger.
Now the theorems in your question do not talk about the fixed fields of individual elements of $\operatorname{Aut}(K/F)$ but about the fixed field of the group $\operatorname{Aut}(K/F)$ as a whole. By that we mean the subfield of $K$ consisting of those elements that are fixed by every automorphism in $\operatorname{Aut}(K/F)$. In other words, it is the intersection of the fixed fields of the individual automorphims. This is again a field, it certainly contains $F$ and we would like it to be exactly $F$. But alas, this last part still need not be true: For example, for the field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ the automorphism group consists only of the identity. Hence the fixed field of $\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$ is $\mathbb{Q}(\sqrt[3]{2})$ and not $\mathbb{Q}$.
Quite soon you should learn that what we want is true for certain field extensions $K/F$ called Galois extensions and these are therefore the main interest of Galois theory.