Yes. $K$ may intersect non-trivially with a cyclotomic extension of $\Bbb{Q}$. The first example that comes to mind is $n=5$, $K=\Bbb{Q}(\sqrt{5})$. Here $K\subseteq\Bbb{Q}(\zeta_5)$. Or more precisely, $K=\Bbb{Q}(\zeta_5)\cap \Bbb{R}$. So $F=K(\zeta_5)$ is only a degree two extension of $K$, and the Galois group is cyclic of order two. The complex conjugation in $F$ is its only non-trivial automorphism, so $Gal(K/F)=\\{\overline{1},\overline{-1}\\}\le\Bbb{Z}_5^*$.