If $G=\mathsf{Aut}_k(F)$ acts on field $F$ algebraic over $k$ then do we have: orbit $G\alpha=\text{ roots of minimal polynomial of }\alpha$? I posed this question on Math.Stackexchange (see here) but until now there was no response. This made me decide to give it a try here.

Let $k\subseteq F$ denote an algebraic field extension and let $\alpha\in F$ having $f\in k[x]$ as its minimal polynomial. Further let $G=\mathsf{Aut}_k(F)$.
My question:

If $\beta\in F$ is a root of $f$ then does there exists some $\sigma\in G$ with $\sigma(\alpha)=\beta$?

In other words: is every root of $f$ in $F$ also an element of orbit $G\alpha$?
I know that the answer is "yes" if the extension is normal but am puzzling whether this condition can be dropped.
Thank you in advance for taking notice of this question, and sorry if it is a duplicate (or for some other reason not suitable for MathOverflow).

Edit: at first hand I forgot to state that $\beta$ is assumed to be an element of $F$. That is repaired now by. Sorry for confusion.
 A: Here is a different interpretation of the question, which is hopefully closer to OP's intent:

Let $F/k$ be an algebraic field extension, and let $\alpha\in F$. Does $Aut_k(F)$ act transitively on the conjugates of $\alpha$ which are contained in $F$?

The answer to this question is no in general. For instance, consider $k=\mathbb Q$, $F=\mathbb Q(\sqrt[4]{2})$, and $\alpha=\sqrt{2}$. Then $F$ contains the conjugate $-\sqrt{2}$ of $\alpha$, but there is no automorphism of $F$ carrying $\sqrt{2}$ to $-\sqrt{2}$, because the former is a square in $F$ and the latter is not.

There is, however, one important case where the answer is positive, specifically when $F=k(\alpha)$. Indeed, in this case, for any conjugate $\beta$ of $\alpha$ contained in $F$, we must have $k(\beta)=F$ as well, since the two have the same degree over $k$ (equal to the degree of the minimal polynomial $\alpha$ and $\beta$). From standard field theory you get an isomorphism from $k(\alpha)$ to $k(\beta)$ fixing $k$ and taking $\alpha$ to $\beta$, which is then an automorphism of $F$.
A: 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?
