Prove that a field automorphism sends a root into a root I came across the following problem:

If $E$ is an extension of $F$ and if $f(x)\in F[x]$ and if $\phi$ is an automorphism of $E$ leaving every element of $F$ fixed, prove that $\phi$ must take a root of $f(x)$ lying in $E$ into a root of $f(x)$ in $E$.

Here's what I have so far:
Let $a\in E$ be a root of $f(x)$. That is, $f(a)=0$. Define an automorphism $\phi: E \to E$ by $\phi(x)=x$. Trivially, the image of any root of $f(x)$ under $\phi$ is a root in the codomain $E$. That proves that the identity automorphism sends a root into a root.
I'm not sure what the other automorphisms are that I must consider. A hint would be greatly appreciated.

You need to consider the case that $\phi$ is an arbitrary automorphism of $E$, such that for every $f \in F$, it is true that
$$\phi(f) = f$$
As a hint for the general case, write down a polynomial with coefficients in $F$; it looks like
$$f_n x^n + f_{n _ 1} x^{n - 1} + \dots + f_1 x + f_0$$
Now apply $\phi$ to this, using the above property.
