Number fields with only trivial field automorphism 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? 

 A: 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.
