Example. [2106A]

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt{ 2 })$. One can check that $$ \tau : \mathbb{Q}(\sqrt{ 2 }) \to \mathbb{Q}(\sqrt{ 2 }), \qquad \tau(a + b\sqrt{ 2 }) = a - b\sqrt{ 2 }, \, a, b \in \mathbb{Q} $$ is an automorphism in $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$. If $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$, then $\sigma(\sqrt{ 2 })^{2} = \sigma(\sqrt{ 2 }^{2}) = \sigma(2) = 2$. Hence, $\sigma(\sqrt{ 2 }) = \pm \sqrt{ 2 }$. If $\sigma(\sqrt{ 2 }) = \sqrt{ 2 }$, then $\sigma = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$, then $\sigma = \tau$. Thus, $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right) = \{ \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}, \tau \} \cong \mathbb{Z} / 2\mathbb{Z}$ ($\tau \circ \tau = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$).