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Fields with trivial automorphism group have been addressed in a question on Mathoverflow (see this). However, I didn't find sufficient information of number fields with such properties.

Q. Let $K$ be a number field whose only field automorphism is trivial one. What properties $K$ must satisfy? For example, can it contain some roots of unity (other than $1,-1$)? Can their degree over $\mathbb{Q}$ be prime power? any composite number?

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    $\begingroup$ If $K/\Bbb Q$ is a number field there is a Galois closure $L/\Bbb Q$ with Galois group $G$ and unique subgroup $H$ for which $K=L^H$ (the elements in $L$ fixed by all elements in $H$). For $g$ to stabilize $K$, it must preserve the property of being fixed by $h\in H$, so $hgk=gk$ for all $h\in H,k\in K$, or equivalently $(g^{-1}hg)k=k$ which means $g^{-1}hg\in H$ so $g$ normalizes $H$. Note every automorphism of $K$ extends to one of $L$, so this means the automorphism group of $K$ is $N_G(H)/H$. To be trivial, $H$ must be its own normalizer. $\endgroup$ – runway44 Jul 1 at 4:40
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    $\begingroup$ Under the Galois correspondence between subfields of $L$ and subgroups of $G$, Galois extensions correspond to normal subgroups and Galois closures correspond to normal cores. Thus, for the Galois closure of $K$ to be $L$, the normal core of $H$ must be trivial. In other words, the Galois group $G$ cannot contain any nontrivial normal subgroup itself contained within $H$. $\endgroup$ – runway44 Jul 1 at 4:40
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You have a wide range of examples. Obviously, extensiosn of degree $2$ have a non trivial automorphisms, so assume $n\geq 3$.

Assume that $K/\mathbb{Q}$ is an extension of degree $n$ which has no non-trivial subextensions and which is not cyclic of prime degree (this always exists, see below) . Then any automorphism of $K$ is trivial.

Indeed,assume that $\sigma$ is an automorphism.

By Artin's lemma, $K/K^{\langle \sigma\rangle}$ has degree $o(\sigma)$.Now $K^{\langle \sigma\rangle}$ is a subextension. By assumption,$K^{\langle \sigma\rangle}=K$, and in this casee $o(\sigma)=1$ and $\sigma=Id$, or $K^{\langle \sigma\rangle}=\mathbb{Q}$. In this case, it means that it is cyclic, and since there is no nontrivial extensions, then $n$ must be prime.

To construct such $K$, here is how to proceed: take the generic extension $\mathbb{Q}(t_1,\ldots,t_n)/\mathbb{Q}(t_1,\ldots,t_n)^{S_n}$. Since $\mathbb{Q}$ is Hilbertian, you can always specialize to a Galois extension $L/\mathbb{Q}$ of group $S_n$. Now set $K=L^{S_{n-1}}$. Since there is no subgroups of $S_n$ containing $S_n$, we are done ($K$ cannot be Galois because $S_{n-1}$ is not normal in $S_n$.)

It proves that any degree $n\geq 3$ is possible.

Notice that you can replace $\mathbb{Q}$ by any number field $F$ to get an extension of degree $n$ without intermediate subfields.

Using this, I think you can construct new families of examples like this:

Take $F/\mathbb{Q}$ and $K/F$ two extensions without intermediate subfields, and which are not cyclic of prime order. Then I suspect that $K/\mathbb{Q}$ has trivial automorphism group, but I have not checked it works.

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  • $\begingroup$ Yes, i forgot the cas of cyclic extensiosn of prime order. I modify my answer right away ! $\endgroup$ – GreginGre Jul 6 at 7:04

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