It seems pretty clear that this statement should be true, but I can't seem to figure out how to prove it.
What I aim to prove is this: Given an irreducible polynomial $p(x)$ in $F[x]$, if $E$ is the splitting field of $p(x)$ in $F$, then for any two zeroes $a_1$ and $a_2$ of $p(x)$ there is an automorphism on $E$ that sends $a_1$ to $a_2$. (*edited due to point made by commentor).
I tried and I think I found a way to prove that if $a_1$ and $a_2$ are two zeroes I can create an automorphism on $F[a_1,a_2]$ that sends $a_1$ to $a_2$. The problem here is there's no guarantee that $a_1$ and $a_2$ split $p(x)$. And I can't seem to extend the automorphism so it works on all of $E$.
This feels like something that should be easy but I just can't seem to get it. This is for homework, but it isn't the full problem (there's probably other ways to do the problem, but my inability to prove this theorem is really bothersome).
Let $\alpha$ and $\beta$ be two roots of $p$ in $E$. Then $F(\alpha) \cong F(\beta)$ as field extensions of $F$. If this isn't obvious, note they are both isomorphic to $F[x] / p(x)$.
The embedding $F(\alpha) \hookrightarrow E$ makes $E$ the splitting field of $p$ over $F(\alpha)$. The embedding $F(\alpha) \to F(\beta) \hookrightarrow E$ also makes $E$ the splitting field of $p$ over $F(\alpha)$. Because the splitting field of a polynomial over a field is unique, these two different extensions of $F(\alpha)$ must be isomorphic. This gives an isomorphism of $E$ that sends $\alpha$ to $\beta$.