It is clear that $\Bbb{Z}_n\otimes_{\Bbb{Z}} \Bbb{Z}_m\cong \Bbb{Z}_{\gcd(m,n)}\cong \text{Hom}_{\Bbb{Z}}(\Bbb{Z}_n,\Bbb{Z}_m)$ and I can construct the morphism directly. I can also swap $\Bbb{Z}_m$ and $\Bbb{Z}_n$'s position if I want.
Q: The question is whether I can obtain this from adjunction formula between $\text{Hom}$ and $\otimes$? The purpose is to ask whether knowing $\Bbb{Z}_n\otimes \Bbb{Z}_m$ is sufficient to determine $\text{Hom}(\Bbb{Z}_n,\Bbb{Z}_m)$ and vice versa. I think this is purely coincidence that $\Bbb{Z}_n\otimes_{\Bbb{Z}} \Bbb{Z}_m\cong \Bbb{Z}_{\gcd(m,n)}\cong \text{Hom}_{\Bbb{Z}}(\Bbb{Z}_n,\Bbb{Z}_m)$.
$\Bbb{Z}_n\cong \text{Hom}(\Bbb{Z},\Bbb{Z}_n)$. However I do not expect to have $\text{Hom}(\Bbb{Z},\Bbb{Z}_n)\otimes \Bbb{Z}_m\cong \text{Hom}(\Bbb{Z}_m,\Bbb{Z}_n)$ but I have $\text{Hom}(\Bbb{Z},\Bbb{Z}_n\otimes \Bbb{Z}_m)$ instead by universal property of tensor product. $\text{Hom}(\Bbb{Z},\Bbb{Z}_n)\otimes \Bbb{Z}_m\cong \text{Hom}(\Bbb{Z}_m,\Bbb{Z}_n)$ may not be even natural in the category of $\Bbb{Z}-$modules.