Lets say $K/F$ is a field extension and $\alpha ,\alpha '\in K$ are two distinct roots of the same irreducible polynomial in $F[x]$. there exists an isomorphism $$\psi:F(\alpha)\rightarrow F(\alpha')$$ which maps $\alpha \mapsto \alpha'$ and fixes the other elements of $F$.
Is there a way to extend this isomorphism to an automorphism $\phi$ of $K/F$ such that $$\phi \rvert_{F(\alpha)}=\psi$$
For example, if $\mathbb{Q}[2^{1/2}]\subseteq \mathbb{R}$, such an extension can't occur for $x^2-2^{1/2}$, sending $2^{1/4}\mapsto -2^{1/4}$. If it did, it would send $2^{1/8}$, a root of $x^2-2^{1/4}$, to a root of $x^2+2^{1/4}$, which doesn't exist