As Robin as pointed out, for all primes $p$, $\mathbb{Q}_p$ is rigid, i.e., has no nontrivial automorphisms. It is sort of a coincidence that you ask, since I spent much of the last $12$ hours writing up some material on multiply complete fields which has applications here:
Theorem (Schmidt): Let $K$ be a field which is complete with respect to two inequivalent nontrivial norms (i.e., the two norms induce distinct nondiscrete topologies). Then $K$ is algebraically closed.
Corollary: Let $K$ be a field which is complete with respect to a nontrivial norm and not algebraically closed. Then every automorphism of $K$ is continuous with respect to the norm topology. (Proof: To say that $\sigma$ is a discontinuous automorphism is to say that the pulled back norm $\sigma^*|| \ ||: x \mapsto ||\sigma(x)||$ is inequivalent to $|| \ ||$. Thus Schmidt's theorem applies.
In particular this applies to show that $\mathbb{Q}_p$ and $\mathbb{R}$ are rigid, since every continuous automorphism is determined by its values on the dense subspace $\mathbb{Q}$, hence the identity is the only possibility. (It is possible to give a much more elementary proof of these facts, e.g. using the Ostrowski classification of absolute values on $\mathbb{Q}$.)
At the other extreme, each algebraically closed field $K$ has the largest conceivable automorphism group: $\# \operatorname{Aut}(K) = 2^{\# K}$: e.g. Theorem 80 of
http://alpha.math.uga.edu/~pete/FieldTheory.pdf.
There is a very nice theorem of Bjorn Poonen which is reminiscent, though does not directly answer, your other question. For any field $K$ whatsoever, and any $g \geq 3$, there exists a genus $g$ function field $K(C)$ over $K$ such that $\operatorname{Aut}(K(C)/K)$ is trivial. However there may be other automorphisms which do not fix $K$ pointwise.
There is also a sense in which for each $d \geq 3$, if you pick a degree $d$ polynomial $P$ with $\mathbb{Q}$-coefficients at random, then with probability $1$ it is irreducible and $\mathbb{Q}[t]/(P)$ is rigid. By Galois theory this happens whenever $P$ is irreducible with Galois group $S_d$, and by Hilbert Irreducibility the complement of this set is small: e.g. it is "thin" in the sense of Serre.
Addendum: Recall also Cassels' embedding theorem (J.W.S. Cassels, An embedding theorem for fields, Bull. Austral. Math. Soc. 14 (1976), 193-198): every finitely generated field of characteristic $0$ can be embedded in $\mathbb{Q}_p$ for infinitely many primes $p$. It would be nice to know some positive characteristic analogue that would allow us to deduce that a finitely generated field of positive characteristic can be embedded in a rigid field (so far as I know it is conceivable that every finitely generated field of positive characteristic can be embedded in some Laurent series field
$\mathbb{F}_q((t))$, but even if this is true it does not have the same consequence, since Laurent series fields certainly have nontrivial automorphisms).
EDIT. Here is the part of the answer that has been rewritten:
We give below a short proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below).
The derivation of the FTGT from (a) and (b) takes about four lines, but I haven't been able to find these four lines in the literature, and all the proofs of the FTGT I have seen so far are much more complicated. So, if you find either a mistake in these four lines, or a trace of them the literature, please let me know.
The argument is essentially taken from Chapter II (link) of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin's proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin's argument) was written with Keith Conrad's help.
Theorem. Let $E/F$ be an extension of fields, let $a_1,\dots,a_n$ be distinct generators of $E/F$ such that the product of the $X-a_i$ is in $F[X]$. Then
the group $G$ of automorphisms of $E/F$ is finite,
there is a bijective correspondence between the sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have
$$
S\leftrightarrow H\iff H=\text{Aut}(E/S)\iff S=E^H
$$
$$
\implies[E:S]=|H|,
$$
where $E^H$ is the fixed subfield of $H$, where $[E:S]$ is the degree (that is the dimension) of $E$ over $S$, and where $|H|$ is the order of $H$.
PROOF
We claim:
(a) If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|\text{Aut}(E/S)|$.
(b) If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.
Proof that (a) and (b) imply the theorem. Let $S/F$ be a sub-extension of $E/F$ and put $H:=\text{Aut}(E/S)$. Then we have trivially $S\subset E^H$, and (a) and (b) imply
$$
[E:S]=[E:E^H].
$$
Conversely let $H$ be a subgroup of $G$ and set $\overline H:=\text{Aut}(E/E^H)$. Then we have trivially $H\subset\overline H$, and (a) and (b) imply $|H|=|\overline H|$.
Proof of (a). Let $1\le i\le n$. Put $K:=S(a_1,\dots,a_{i-1})$ and $L:=K(a_i)$. It suffices to check that any $F$-embedding $\phi$ of $K$ in $E$ has exactly $[L:K]$ extensions to an $F$-embedding $\Phi$ of $L$ in $E$; or, equivalently, that the polynomial $p\in\phi(K)[X]$ which is the image under $\phi$ of the minimal polynomial of $a_i$ over $K$ has $[L:K]$ distinct roots in $E$. But this is clear since $p$ divides the product of the $X-a_j$.
Proof of (b). In view of (a) it is enough to check $|H|\ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a
$$
b=(b_1,\dots,b_k)\in E^k.
$$
We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^\perp\cap(E^H)^k$ is nonzero, where $\bullet^\perp$ denotes the vectors orthogonal to $\bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $b^\perp\cap (E^H)^k$ is necessarily orthogonal to $hb$ for any $h\in H$, so
$$
b^\perp\cap(E^H)^k=(Hb)^\perp\cap(E^H)^k,
$$
where $Hb$ is the $H$-orbit of $b$. We will show $(Hb)^\perp\cap(E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^\perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^\perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^\perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j=1$ for some $j$. Since the subspace $(Hb)^\perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $hx\in(Hb)^\perp$, so $hx-x\in(Hb)^\perp$. Since $x_j=1$, the $j$-th coordinate of $hx-x$ is $0$, so $hx-x=0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.
[A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II, available here.
PDF version: http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Selected_Texts/st.pdf
Here is the part of the answer that has not been rewritten:
Although I'm very interested in the history of Galois Theory, I know almost nothing about it. Here are a few things I believe. Thank you for correcting me if I'm wrong. My main source is
http://www-history.mcs.st-and.ac.uk/history/Projects/Brunk/Chapters/Ch3.html
Artin was the first mathematician to formulate Galois Theory in terms of a lattice anti-isomorphism.
The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.
The first publications of this formulation by Artin himself were "Foundations of Galois Theory" (1938) and "Galois Theory" (1942).
Artin himself doesn't seem to have ever explicitly claimed this discovery.
Assuming all this is true, my (probably naive) question is:
Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?
I also hope this is not completely unrelated to the question.
Best Answer
Even in our day of sophisticated search engines, it still seems that the success of a search often turns on knowing exactly the right keyword.
I just followed up on Sylvain Bonnot's comment above. The property of a field extension $K/F$ that for all subextensions $L$ we have $K^{\operatorname{Aut}(K/L)} = L$ is apparently most commonly called Dedekind. This terminology appears in Exercise V.9 of Bourbaki's Algebra II, where the reader is asked to show that if $L/K$ is a nonalgebraic Dedekind extension and $T$ is a transcendence basis, then $L/K(T)$ must have infinite degree. Ironically, this is exactly what I could show in my note. One can (in the general case, even...) immediately reduce to the case $T = \{t\}$ and then the exercise is saying that the function field $K(C)$ of an algebraic curve (again, it is no loss of generality to assume the function field is regular by enlarging $K$) is not Dedekind over $K$. This is kind of a strange coincidence! [However, the proof I give is openly geometric so is probably not the one that N.B. had in mind...]
It also appears in
In this paper, the author shows that $L/K$ is a Dedekind extension iff for all subextensions $M$, the algebraic closure $M^*$ of $M$ in $L$ is such that $M^*/M$ is Galois in the usual sense: i.e., normal and separable. (This is a nice fact, I suppose, and I didn't know it before, but it seems that the author regarded this as a solution of the problem of which extensions are Dedekind. I don't agree with that, since it doesn't answer my question!)
Apparently one is not supposed to read the above paper but rather this one:
Here is the MathSciNet review by E.R. Kolchin (who knew something about transcendental Galois extensions!):
So it seems that my question is a nearly 60 year-old problem which was considered but left unsolved by Krull. I am tempted to officially give up at this point, and perhaps write up an expository note informing (and warning?) contemporary readers about this circle of ideas. Comments, suggestions and/or advice would be most welcome...
P.S.: Thanks very much to M. Bonnot.