What is the number of distinct homomorphism from $\Bbb Z/5 \Bbb Z$ to $\Bbb Z/7 \Bbb Z$ 
What is the number of distinct homomorphism from $\Bbb Z/5 \Bbb Z$  to  $\Bbb Z/7 \Bbb Z$ and how to find it?  

I came across the above problem and do not know how to get it? Can someone point me in the right direction? Thanks in advance for your time.

Hint 1 (looking at subgroups and their orders): Any homomorphism $\Bbb{Z}_5\xrightarrow{\phi}\Bbb{Z}_7$ yields a subgroup of $\Bbb{Z}_7$ given by $\phi\left(\Bbb{Z}_5\right)$. What are the subgroups of $\Bbb{Z}_7$? Given the subgroups and their orders, what possibilites are there for $\phi\left(\Bbb{Z}_5\right)$? (You might want to use Lagrange's theorem.)
Hint 2 (following the hint in the comment): Both groups are cyclic and generated by $1$. Therefore, a homomorphism $\phi : \Bbb{Z}_5\to\Bbb{Z}_7$ is determined by where $1$ is sent (since any element $k\in\Bbb{Z}_5$ is equal to $k\cdot 1 = \underbrace{1 + \ldots + 1}_{k\textrm{ times}}$). So $\phi(1)$ generates a subgroup of $\Bbb{Z}_7$. Think about the possibilities for the size of the subgroup given by $\left<\phi(1)\right>\subseteq\Bbb{Z}_7$ based on the orders of $\Bbb{Z}_5$ and $\Bbb{Z}_7$ (and hence the possibilities for $\phi(1)$).
