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 $\beta=\alpha$.)

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 $\beta=\alpha$.)

In view of the persistent criticism by YCor below, it's perhaps worth stating very explicitly the question I'm answering here (admittedly, quite trivially) and that I think the OP asked:

Suppose that $F/k$ is an algebraic field extension with the following property: if $\alpha,\beta\in F$ are zeros of an irreducible $f\in k[x]$, then there is a $k$-automorphism of $F$ mapping $\alpha\mapsto\beta$. Does it follow that $F/k$ is normal?

Yes, this can hold for non-normal extensions, for example $F=\mathbb Q(2^{1/3})$. 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, if $\alpha\notin\mathbb Q$, its minimal polynomial will have no other zeros in $F$.)

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 $\beta=\alpha$.)