Originally Posted by
NonCommAlg well, the automorphism group of F is actually a cyclic group of order exactly n, but i guess you don't want to get there yet. to answer your question:
suppose $\displaystyle \alpha$ is an element of F whose minimal polynomial over $\displaystyle K=\mathbb{Z}/p$ has degree $\displaystyle n$. call this polynomial $\displaystyle f(x).$ then $\displaystyle F=K(\alpha).$ now any $\displaystyle \sigma \in \text{Aut}(F)$ fixes elements of $\displaystyle K$ and so it's completely
determined by $\displaystyle \sigma(\alpha)$ and $\displaystyle f(\sigma(\alpha))=\sigma(f(\alpha))=\sigma(0)= 0.$ so there are at most $\displaystyle n$ possibilities for $\displaystyle \sigma(\alpha).$