Aut($C_p) = C_{p-1}$when $p$ is prime, why? Aut($C_p) = C_{p-1}$ when $p$ is prime, why?
I don't see why this result follows, I'm sure it's obvious though.
I appreciate that Aut($C_p)$ will be of order p-1
 A: Pick a fixed generating element $e$. Every endomorphism $f$ is uniquely determined by the image of this generating element $f(e)$. 
The image of the generating element $f(e)$ must  be generating to have an automorphism. 
There are thus $p-1$ possibilities for $f(e)$; every element but the neutral one.
To show that the group is cyclic it is best to consider $C_p$ as a field. 
Derek Holt already mentioned a general result that would imply this. 
To prove it denote by $t$ the maximal multiplicative order; each non-zero elements multiplicative order is a divisor of it. So each nonzero element is a solution of $X^t=1$. Yet since $X^t-1$ can have at most $t$ roots, we get $t=p-1$, and thus an element of order $p-1$.
A: Write $C_p=\{e,x,x^2,...,x^{p-1}\}$ where $e$ is the identity element. An automorphism is by definition an isomorphism and an endomorphism, which can be thought of as a relableing of $C_p$ such that $f(xy)=f(x)f(y)$ (where $f$ is any given automorphism in question). Let's construct a specific automorphism as follows: Let $f$ be an automorphism. Then by properties of homomorphisms, $f(e)=e$. Now assume $f(x)=x^a$ where $a$ is a primitive root. Then $f(x^n)=f(x)^n=(x^a)^n=x^{an}$. Then by composing $f$ with itself, we get that since $a$ is a primitive root, for all $0<n<p$ there is an $0<m<p-1$, $(f^m)(x)=x^n$, hence this process gives all automorphisms and we can check that all of these compositions of $f$ are automorphisms.
