Finding the fixed subfield corresponding to a cyclic subgroup of the Galois group

Let's say I have a field extension $E$ of some field $F$ and I also know the Galois group of $E$ over $F$. Suppose I have a subset of this Galois group which is cyclic, thus generated by some automorphism $\alpha$. When looking for the subfield of $E$ fixed by this subgroup, I know it's best to see what the automorphism does to a typical element of $E$ (let's call it $x$) which would be a linear combination of the basis elements of $E$. My question is, would it be sufficient to check the action of $\alpha$ on $x$ to determine the fixed subfield or would I also have to check the action of $\alpha^2, \alpha^3$ and other elements of the cyclic subgroup?

You know, if $\alpha(x)=x$, then $\alpha^2(x)=\alpha(\alpha(x))=\alpha(x)=x$ and use induction to show that $\alpha^n(x)=x$. So it's enough to check $\alpha$.