A quantity associated to a field extension Let $F\subset E$ be a field extension. So $E$ has a natural structure of $F$-vector space.
A vector subspace $V\subset E$ is a special subspace if $F\subset V$ and $V$ is closed under the inverse operator $i \colon E\to E$ with $i(a)=a^{-1}$.

Question. For what kind of field extensions $F\subset E$ do we have but finitely many special subspaces? Is there a well-known
interpretation for the number of such special subspaces when $E$ is a
finite field?

 A: For the rest of this post, I'm talking about fields not of characteristic $2$. Under this assumption, we will show that any such $V$ is either a subfield of $E$ or is a complement of a subfield $F'$ inside a quadratic extension $F' \oplus V$ of $F'$ of the form $F' \sqrt{\alpha}$ for some $\alpha \in F'$.
We first prove that if $a, b \in V$ with $b\neq 0$, then $\frac{a^2}{b} \in V$. I will list expressions that must also be elements of $V$:
$\frac{1}{a}, \frac{1}{b}, \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}, \frac{ab}{a + b}, \frac{ab}{a + b} - a = -\frac{a^2}{a + b}, \frac{a + b}{a^2}, \frac{a + b}{a^2} - \frac{1}{a} = \frac{b}{a^2}, \frac{a^2}{b}$
(Here we only run into problems with inversion at $0$ if one of $a$, $b$, or $a + b$ is zero; if $a=0$ then $\frac{a^2}{b}=0 \in V$, while if $a+b=0$ then $\frac{a^2}{b}= \frac{b^2}{b} =b\in V$, so the claim is valid even in these cases.)
We next prove that if $a, b, c \in V$ with $c\neq 0$, then $\frac{ab}{c} \in V$. Again, I will list expressions that must be elements of $V$:
$\frac{(a + b)^2}{c}, \frac{a^2 + b^2}{c}, \frac{2ab}{c}, \frac{ab}{c}$.
Fix some $c\in V$ nonzero. Let $F'$ be the set of elements of $E$ of the form $\frac{b}{c}$ for $b\in V$. Then I claim that $F'$ is closed under addition, multiplication, and inversion, and so is a field. Addition is automatic, multiplication follows from $\frac{a}{c} \frac{b}{c} = \left(\frac{ab}{c} \right)/c$, and inversion follows from $\left(\frac{b}{c}\right)^{-1} = \left( \frac{c^2}{b} \right)/c$.
But then we have that $c^{2} = \left( \frac{c^2}{c^{-1}}\right)/c \in F'$, so $F' \oplus V = F'[c]$ is a quadratic extension of $F'$ (unless $V = F'$ itself). So what you're looking for really are pairs of subfields of $E$ where one is a quadratic extension of the other, or just subfields of $E$.

We now turn to the characteristic 2 case. For $E$ finite, the answer is still simple: $V$ must be a subfield of $E$.
Proof: First, I claim that if $V$ is nontrivial, then $1 \in V$.
Choose any nonzero $c \in V$, and choose some generator $\alpha \in E$. Let $n$ be the smallest nonzero exponent such that $c \alpha^n \in V$. Then we can use induction to show that $c \alpha^{xn} \in V$ for $x \in \mathbb{Z}$, as $c \alpha^{(x + 1) n} = \frac{(c \alpha^{xn})^2}{c \alpha^{(x - 1)n}}$, and similar induction for negative exponents. In the other direction, we can see that if $d \in V$, then as $\alpha$ is a generator, we have that $d = c \alpha^m$ for some $m \in \mathbb{N}$; it's not hard to see that $n|m$ (as otherwise, $n$ wouldn't be the smallest), so we have that $V = \{c \alpha^{xn}\}_{x \in \mathbb{N}} \cup \{0\}$.
Note that $|E| - 1$ must be divisible by $n$ - and in particular, $n$ is odd. But then as $c^{-1} \in V$, we must have that $c^{-1} = c \alpha^{x n}$, so $c^{2} = \alpha^{(|E| - 1) - xn} = \alpha^{2(|E| - 1) - xn}$. One of the two exponents is even, and half of that is still divisible by $n$ - so we get that $c = \alpha^{x_0 n}$ for some $x_0$. But then $1 = \alpha^{|E| - 1} = c \alpha^{|E| - 1 - x_0 n} \in V$.
By applying the above argument, this time with $c = 1$, we get that $V = \alpha^{xn} \cup \{0\}$ - so it must be closed under multiplication. We already had that it was closed under inversion and addition, so it is a subfield of $E$.
The question is still open for $E$ infinite of characteristic 2.
A: For $F$ of characteristic $2$, a space $V$ has this form if and only if there are fields $F_1, F_2$ with $F \subseteq F_1 \subseteq F_2 \subseteq E$ and $F_2^2 \subseteq F_1$, and $V$ is an $F_1$-linear subspace of $F_2$.
For "if", by definition $V$ is $F_1$-linear, thus $F$-linear, and if $0 \neq a \in V$ then $a \in F_2$ so $a^2 \in F_1$ and thus $a^{-1} =(a^2)^{-1} a  \in V$ by $F_1$-linearity.
For "only if", by user44191's answer, if $a, b\in V$ then $a^2 b^{-1} \in V$. Because $V$ is closed under inverses, we get $a^2 b \in V$. Let $F_1$ be the field generated by squares of elements of $V$, then this identity shows $V$ is stable under multiplication by elements of $F_1$. Let $F_2$ be the field generated by $V$. We have $F_1 \subseteq F_2$ and $V$ is an $F_1$-linear subspace of $F_2$. Finally, because squaring is a field homomorphism in characteristic $2$ and $V^2 \subseteq F_1$ by definition, $F_2^2 \subseteq F_1$.
Thus, in characteristic $2$, there are finitely many such if and only if $E/F$ is finite separable — otherwise we can find a nontrivial pair $F_1$, $F_2$ and then there will be infinitely many one-dimensional $F_1$-linear subspaces of $F_2$ (we can assume $F_1$ is infinite because finite fields have no inseparable extensions).
