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Yes, this can hold for non-normal extensions, for example $F=\mathbb Q(2^{1/3})$. For any zeros $\alpha,\beta\in F$ of an irreducible $f\in\mathbb Q[x]$, we can always map $\alpha\mapsto\beta$ by a $\mathbb Q$-automorphism $\sigma: \mathbb Q(\alpha)\to \mathbb Q(\beta)$, and there are no intermediate fields here since $[F:\mathbb Q]=3$. (So in the example, it follows that $\beta=\alpha$.)