Describe explicitly $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_n, \mathbb{Z}\_m) := \{\varphi:\mathbb{Z}\_n \rightarrow \mathbb{Z}\_m \mid \mathbb{Z}\text{-linear homomorphism}\}$

There is an answer to this question that says $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_n, \mathbb{Z}\_m) \cong \mathbb{Z}\_{(n,m)}$ where $(n,m)=\gcd(n,m)$. But I am having some trouble seeing this.

For the sake of an argument, suppose the above answer is true, that $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_n, \mathbb{Z}\_m) \cong \mathbb{Z}\_{(n,m)}$. Suppose $n=4$ and $m=8$. Then $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_4, \mathbb{Z}\_8) \cong \mathbb{Z}\_4$. 

Now consider each of the following maps in $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_4, \mathbb{Z}\_8)$:

$\varphi\_1(1)=1$ 

$\varphi\_2(1)=2$ 

$\varphi\_3(1)=3$ 

$\varphi\_4(1)=4$ 

$\varphi\_5(1)=5$

$\varphi\_6(1)=6$

$\varphi\_7(1)=7$

Are each of these maps not unique? If there are two that are identical, could you please explain which two and why? Thank you!

My alternate answer is that $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_n, \mathbb{Z}\_m) \cong \mathbb{Z}\_m$, since I think each $\varphi \in \operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}\_n, \mathbb{Z}\_m)$ is uniquely determined by where it maps $\varphi(1) \in \mathbb{Z}\_m$.