Morphisms between cyclic groups  I'm trying to solve a group theory question involving morphisms: 
How many different morphisms do there exist from $ C_n $ to $ C_m $? 

Am I correct in saying if $f$ is a morphism, then $f(0) = 0$ and $f(a+b) = (a+b)f(1)$? 
If yes, where do I go from here? And if no, how do I start? 
Thanks! 

Also, as a follow up question there's: 
How many automorphisms exist from $C_n$ onto itself? 
Since automorphisms map generators to generators, I have to map $1$ to a generator, so I have $\phi(n)$ choices, with $\phi$ being Euler's totient function?
 A: You are right in your assumption (the phrase used by mathematicians is "A homomorphism is uniquely determined by its image on the generators"), and from there you just have to make sure that the image of the generators behave like they originally do; in this case the one generator $1 \in C_n$ fulfills $n\cdot 1 = 0$
By your assumtion, the homomorphism $\psi$ is completely determined by $\psi(1)$. So the question reads "Where can I send $1$?", and the only limiting factor here is that in $C_n$, $n\cdot 1 = 0$, so we must have $n\cdot \psi(1) = 0$ in $C_m$. Now the question reads "Which elements in $C_m$ has order a divisor of $n$?". This is when we leave algebra and enter number theory, where the same question sounds "Which integers $k$ fulfills $k\cdot n \equiv 0 \mod{m}$?" And the answer is any multiple of $\frac{m}{\gcd(m, n)}$, of which there are $\gcd(m, n)$, including $0$.
As to your follow-up, you can map $1$ to any element $j$ with $\gcd(n, j) = 1$, of which there are $\phi(n)$. If you map it to any other element, it will not generate a surjection, since $1$ would not be in the image, and thus not an isomorphism. So you are right.
