Is there a field extension $K / \Bbb Q$ such that $\text{Aut}_{\Bbb Q}(K) \cong \Bbb Z$? I'm not requiring this extension to be Galois, that's why I wrote $\text{Aut}$ instead of $\text{Gal}$. I'm not very familiar with infinite extensions nor with profinite groups. I don't know if my question is part of of the inverse Galois problem, since this problem is generally related to finite groups.
To start with the "basic" infinite extension $\Bbb Q(X)$, I know that $\text{Aut}_{\Bbb Q}(\Bbb Q(X)) \cong \text{PGL}_2(\Bbb Q)$, which is far from being $\Bbb Z$. Notice that this question is not really related to mine.
Thanks for your help!
 A: This question is answered affirmatively in the paper
Kuyk, Willem
The construction of fields with infinite cyclic automorphism group.
Canad. J. Math. 17 1965 665-668. 
Armand Brumer wrote this in his MathSciNet review of the paper:
``J. de Groot [Math. Ann. 138 (1959), 80-102, 101; MR0119193] has asked whether every group $G$ is the full automorphism group of some field. The author answers this question affirmatively for the infinite cyclic group. In fact, let $\mathbb K$ be an algebraic closure of a purely transcendental extension $\mathbb Q(t_0)$ of the rationals. Choose inductively elements $t_i$ in $\mathbb K$ satisfying $t_i^2=t_{i−1}+1$ for all integers $i$, and set $\Omega = \bigcup_{i\in \mathbb Z} \mathbb Q(t_i)$. Then the automorphism group of $\Omega$ is the infinite cyclic group generated by $t_i\mapsto t_{i+1} (i\in\mathbb Z)$.''
Later Kuyk's result was extended in:
Fried, E.; Kollár, J.
Automorphism groups of fields. Universal algebra (Esztergom, 1977), pp. 293-303,
Colloq. Math. Soc. János Bolyai, 29, North-Holland, Amsterdam-New York, 1982.
These authors show that for any group $G$ there exists a field $\mathbb F$ such that $\textrm{Aut}(\mathbb F)\cong G$. Moreover, $\mathbb F$ may be required to have any given characteristic different from $2$. 
