Finding Number of Automorphisms of Z8? I'm trying to find the number of automorphisms of Z8.  When I google around, I find stuff like: 
There are 4 since 1 can be carried into any of the 4 generators.
The problem hint tells me to make use of the fact that, if G is a cyclic group with generator a and f: G-->G' is an isomorphism, we know that f(x) is completely determined by f(a).
Thing is, I can think of 7! 1-1 and onto mappings of Z8 onto itself.  I guess I don't see exactly why 1 has to get carried into a generator...why can't I have f(n) = n + 1 (mod 8), just shifting each element one to the right?
Thanks for any guidance on this,
Mariogs
 A: An automorphism is not just one-to-one and onto. It is also a group homomorphism. While there are many one-to-one and onto maps between a set and itself, not all of such maps will preserve the structure of the group. To be a group homomorphism, you must have $f(a+b) = f(a)+f(b)$ for any $a,b\in\mathbb{Z}_8$. Under your shift right suggestion we would get that $f(1+1) = f(2) = 3$ but $f(1)+f(1) = 2+2 = 4$. So it is not a group homomorphism.
A: Note that $1$ generates $\mathbb{Z}_{8}$ as a group, so any group morphism $\varphi:\mathbb{Z}_8\rightarrow\mathbb{Z}_8$ is determined by $\varphi(1)$. Furthermore, if $\varphi$ is an automorphism, then $\varphi(1)$ generates $\mathbb{Z}_8$. The possible generators of $\mathbb{Z}_8$ are $1,3,5,7$. It then remains to check that for each possible choice of generator, there exists $\varphi$ with $\varphi(1)$ equal to the generator. This is the case, so there are $4$ possible automorphisms of $\mathbb{Z}_8$. (To see this, define $\varphi(n)=3n, 5n, 7n$ and check each of these.)
