The number of automorphisms of a finite field Let $M$ be a finite field and $|M| = p^s$, where $p$ is prime and $s \in \mathbb N$. Prove that the number of different isomorphisms field $M$ to $M$ equal to $s$ and this isomorphisms form a cyclic group under composition of isomorphisms.
We know that the multiplicative group of finite field is cyclic and all cyclic groups the same order are isomorphic. Because of this if I do not mistake the number of different isomorphisms $M^{\times}$ to $M^{\times}$ equal to the number of primitive elements in $M^{\times}$. And it is equal to $\phi(p^s - 1)$ where $\phi(n)$ - Euler function. But we should consider not only $M^{\times}$ but also the addition and I do not know what to do with it and how to connect it with the multiplicative group.
Thanks for the help!

To make the question answered I copied for you the relevant paragraph (see pages 208, 209, 210, 211) of a Russian translation of “Algebra” by Serge Lang.  Theorem 12 at p. 211 answers your question.    
