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!
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 $2$.