Definition. Let $F$ be a field and $p(x)$ be a polynomial over $F$. We say that an extension $K:F$ is a minimal splitter for $p(x)$ over $F$ if $K$ splits $p(x)$ and whenever there is an extension $L:F$ of $F$ in which $p(x)$ splits, there is an $F$-embedding $\varphi:K\to L$.
I think that the minimal splitter as defined above is same as the splitting field for $p(x)$ over $F$. If this is indeed true, then the following should hold
Conjecture. Let $p(x)\in F[x]$ be irreducible over $F$ and $K:F$ be a minimal splitter of $p(x)$ over $F$. Let $\alpha$ and $\beta$ be two roots of $p(x)$ in $K$. Then there is an $F$-automorphism of $K$ which sends $\alpha$ to $\beta$.
The roadblock in proving the above is showing that $K:F(\alpha)$ is also a minimal splitter for $p(x)$ over $F(\alpha)$. I know that there is an $F$-isomorphism $F(\alpha)\to F(\beta)$ which sends $\alpha$ to $\beta$. But I am not able to extend this to an $F$-automorphism of $K$.
EDIT:
I think I have got it. Let $K:F$ be the minimal splitter for a polynomial $p(x)$ over $F$. Let $\alpha_1, \ldots, \alpha_n$ be all the roots of $p(x)$ in $K$. Define $E=F(\alpha_1, \ldots, \alpha_n)$. Then $E$ also splits $p(x)$ and therefore we have an $F$-embedding $K\to E$. By comparing dimensions of $K$ and $E$ over $F$ we must have that this embedding is in fact surjective and hence an isomorphism.