I have seen the next statement but using that $K / F$ is normal instead of algebraic, but I think that it can be shown only considering $K / F$ as an algebraic extension of fields.
Let $K / F$ be an algebraic extension of fields and let $x , y \in K$ such that $p(x , F)(X) = p(y , F)(X)$. Then there exists a $F$-automorphism $\sigma : K \to K$ such that $\sigma(x) = y$.
Notes:
- $f(X) = p(x , F)$(X) is the monic and irreducible polynomial in $F[X]$ such that $f(x) = 0$.
- $\sigma : K \to K$ is $F$-automorphism if it is automorphism and $\sigma(a) = a$ for all $a \in F$ ($\sigma$ will be $F$-immersion if it is immersion and $\sigma(a) = a$ for all $a \in F$).
- If $g(X) = \sum_{i = 0}^n a_i X^i$ is a polynomial in $F[X]$, then $g^{\sigma}(X) = \sum_{i = 0}^n \sigma(a_i) X^i$.
Proof: If we consider the inclusion $i : F \hookrightarrow K$, which is a $F$-immersion, then there exists an extension $\sigma : K \to K$ of $i$ ($\sigma(a) = i(a) = a$ for all $a \in F$ and $\sigma$ is a immersion, so $\sigma$ is automatically a $F$-immersion) such that $\sigma(x) = y$ (it happens in all the algebraic extensions), as $y$ is root of $p(x , F) = p(y , F)$. Since the extension is algebraic (here all the $F$-immersions are $F$-automorphisms), $\sigma$ is then a $F$-automorphism.
In the proof, I have used the next statement and honestly I am not sure if it is true:
Let $K / F$ be an algebraic extension of fields and let $L$ be a field such that there exist an immersion $\sigma : F \to L$. Let $x \in K$ and $y \in L$ and we suppose that $y$ is root of the polynomial $p^{\sigma}(x , F)(X)$. Then there exist an extension $\tau : K \to L$ of $\sigma$ ($\tau$ is immersion and $\tau(a) = \sigma(a)$ for all $a \in F$) such that $\tau(x) = y$.