Is there a field extension $K / \Bbb Q$ such that $\text{Aut}_{\Bbb Q}(K) \cong \Bbb Z$?

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 $2$.