Is there a finite field extension with „less“ intermediate fields than Aut-Subgroups? When looking for a counterexample to the „lattice isomorphism“ part of the fundamental theorem of Galois theory, I usually look for non-Galois extensions $K\colon F$ e.g. by choosing $F$ to not contain all conjugates of some element.  Consider for example $$
\mathbb Q(\sqrt[3]{2})\colon \mathbb Q
$$
which has trivial automorphism group, but a two-element lattice of intermediate fields.

Are there examples (in char. 0, since I don't understand anything else) where we have more subgroups of $\operatorname{Aut}(K\vert F)$ than there are intermediate fields?

…Or is the map $$
\varphi\colon \operatorname{Sub}(\operatorname{Aut}(K\vert F)) \to \{E\mid F\leq E \leq K\} \\
 H \mapsto \operatorname{Fix}(H)
$$ injective even for non-Galois (but still finite) extensions?
 A: No- the map is indeed injective (at least if you're dealing with fields of characteristic $0$).
First, some notation. For any subgroup, $H$, of $\operatorname{Aut}(K\mid F)$, denote the subset of all elements of $K$ fixed by every member of $H$, $H^F$ (this is just $\operatorname{Fix}(H)$). For any field intermediate $K$ and $F$, $L$, denote the subset of all elements of $\operatorname{Aut}(K\mid F)$ that fixes every element of $L$, $L^*$.
It is elementary to show that, for any subgroup, $H$, of $\operatorname{Aut}(K\mid F)$, $H^F$ is a field between $F$ and $K$, and that, for any field, $L$, between $F$ and $K$, $L^*$ is a subgroup of $\operatorname{Aut}(K\mid F)$.
The map from $\operatorname{Sub}(\operatorname{Aut}(K\mid F))$ to $\{E\mid E=\mathrm{field}, F\leq E\leq K\}$ under consideration here is the function that sends a subgroup, $H$, of $\operatorname{Aut}(K\mid F)$ to $H^F$. When $\operatorname{char}(F)=0$, it is injective.
In what follows, $H$ represents an arbitrary subgroup of $\operatorname{Aut}(K\mid F)$, and $L$, an arbitrary field between $F$ and $K$.
The first result is that: if $H=L^*$, then $L\subset H^F$.
The proof of this is pretty obvious. If $H=L^*$, then, by definition, every member of $H$ fixes all of $L$. $H^F$ is defined to be all the stuff in $K$ fixed by every member of $H$, of which, $L$ is clearly a part.
The second result is: if $L=H^F$, then $|H|=[K:L]$.
This one is more involved. First, we note that, as $H$ consists solely of $L$-fixing automorphisms on $K$, we must have $|H|\leq [K:L]$. Now, we use a rather complicated argument to show $|H|\geq [K:L]$. Combined with the first inequality, this will yield $|H|=[K:L]$, as desired.
Now, let the elements of $H$ be $h_1,h_2,\dots,h_r$, where $r=|H|$.
Since $L\supset F$ and $\operatorname{char}(F)=0$, we have $\operatorname{char}(L)=0$; we may then write $K=L(a)$ for some $a\in K$.
Consider the polynomial $A(x)=\prod_{i=1}^r(x-h_i(a))$
We note here that, since the identity automorphism appears as some $h_i$, $(x-a)$ is a factor of $A(x)$, i.e. $a$ is a root of $A(x)$.
For each $i$, define a function $\bar h_i$ on $K[X]$ so that, for any $k(x)=k_nx^n+\dots+k_1x+k_0\in K[X]$,
$\bar h_i(k(x))=h_i(k_n)x^n+\dots+h_i(k_1)x+h_i(k_0)$
It is easy to show that each $\bar h_i$ is an automorphism on $K[X]$.
Denote the group of these automorphisms on $K[X]$ by $\overline{H}$.
Now, let us apply an arbitrary $\bar h_j$ to $A(x)$: $$\bar h_j(A(x))=\bar h_j\left(\prod_{i=1}^r(x-h_i(a))\right)$$
As $h_j$ is an automorphism, we have $$\bar h_j(A(x))=\prod_{i=1}^r\bar h_j(x-h_i(a))$$
Using the definition of $\bar h_j$: $$\bar h_j(A(x))=\prod_{i=1}^r(x-h_j(h_i(a)))=\prod_{i=1}^r(x-(h_j\circ h_i)(a))$$
I have used $\circ$ to denote the composition of the automorphisms. Now, we note that, as $h_j$ is an element of $H$, the left coset of $h_j$ with $\overline{H}$, i.e. $h_j\circ H$, is simply equal to $H$. So the set $\{(h_j\circ h_1)(a),(h_j\circ h_2)(a),\dots,(h_j\circ h_r)(a)\}$ again equals $\{h_1(a),h_2(a),\dots,h_r(a)\}$.
Which means $\bar h_j(A(x))=A(x)$. Since $j$ was arbitrary this holds for any element of $\overline{H}$. But, given the definition of $\bar h_j$, for us to have $\bar h_j(A(x))=A(x)$, we must have that $h_j$ fixes every coefficient of $A(x)$. This means every member of $H$ fixes every coefficient of $A(x)$. Since we assumed $L=H^F$, that means $A(x)\in L[x]$.
Recall, however, that $a$ is a root of $A(x)$. This means the minimal polynomial of $a$ over $L$ divides $A(x)$. The degree of the minimal polynomial of $a$ over $L$ is $[L(a):L]=[K:L]$. Since this polynomial divides $A(x)$, we must have $\mathrm{deg}(A(x))\geq [K:L]$
But, by the way $A(x)$ was defined, $\mathrm{deg}(A(x))=|H|$, so we have $|H|\geq [K:L]$, completing the proof of the second result.
The final result is: if $L=H^F$, then $L^*=H$. If we can show this, we'll have that the map from subgroups to intermediate fields is injective. So, suppose we have $L^*=E$. Well, looking back at the definitions, $E$ is the biggest subgroup that fixes all of $L$. $L$ is the field fixed by all of $H$. $H$ fixes all of $L$, so we must have $H\subset E$.
By the $2^{nd}$ result, we also have $|H|=[K:L]$.
By the $1^{st}$ result, we have that $L\subset E^F$, and, by the $2^{nd}$ result, we have $|E|=[K:E^F]$.
Since $L\subset E^F$, we must have $|E|=[K:E^F]\leq[K:L]$, but $|H|=[K:L]$. So $|E|\leq |H|$. Combining this with the fact that $H\subset E$, we have $H=E=L^*$, as desired.
So there will always be more intermediate fields than subgroups.
(If there are any doubts, you can also explicitly show that the map fails to be surjective. Specifically, $F$ will never be $H^F$ for any subgroup, $H$, of $\operatorname{Aut}(K\mid F)$ if $K$ is non-Galois. Why? Suppose not. Suppose it was. What does result $2$ tell you then? Is this possible?)
(please comment/ edit for any corrections)
