If I understand the comment, we are to ignore the whole Galois group thing, and just talk about permuting the coefficients of $p(x)$, and whether the resulting polynomial $q(x)$ can have any common roots with $p(x)$. If $p(x)$ is irreducible then it can't have any common roots with $q(x)$ unless it equals $q(x)$. If $p(x)$ is not irreducible, well, $x^2+3x+2$ and $2x^2+3x+1$ share a root, it's not hard to construct other examples, it's not clear there's anything useful to be said about the situation. Permuting coefficients isn't "natural".