I want to find an explicit description of the group $\mathbf{Hom}(\mathbb{Z}_p^{\times};\mathbb{Z}/n)$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers.
If $p\geq 3$, we have that $\mathbb{Z}_p^{\times}\cong \mathbb{Z}/(p-1)\oplus\mathbb{Z}_p$.
If $n=p^m$, we have that $\mathbf{Hom}(\mathbb{Z}_p;\mathbb{Z}/p^m)\cong\mathbb{Z}/p^m$. Indeed, the short exact sequence $\mathbb{Z}_p \to \mathbb{Z}_p \to\mathbb{Z}/p^m$ (where the first map is the multiplication by $p^m$) induces the following exact sequence $$0\to\mathbb{Z}_p\cong\mathbf{Hom}(\mathbb{Z}_p;\mathbb{Z}_p) \to \mathbf{Hom}(\mathbb{Z}_p;\mathbb{Z}_p) \to \mathbf{Hom}(\mathbb{Z}_p;\mathbb{Z}/p^m)\to 0$$ What about the case of general $n$?
Moreover, what does it happen in the case of $\mathbb{Z}_2^{\times}$?
For any abelian group $(A, +)$, we have obviously $Hom (A, \mathbf Z /n) \cong Hom (A/n, \mathbf Z /n)$, so in particular $Hom({\mathbf Z_p}^*, \mathbf Z /n) \cong Hom({\mathbf Z_p}^*/({\mathbf Z_p}^*)^n, \mathbf Z /n)$. But the structure of ${\mathbf Z_p}^*$ is known : it is $\cong (\mathbf Z_p , +) \times (\mathbf Z /(p-1), +)$ if $p$ is odd, $\cong (\mathbf Z_2 , +) \times (\mathbf Z /2, +)$ if $p=2$. I'll detail only the case $p \neq 2$, writing $(\mathbf F_p^*, \times)$ instead of $(\mathbf Z /(p-1), +)$. The only non obvious point is the determination of $\mathbf F_p^*/(\mathbf F_p^*)^n$. The $n$-th power map $\nu$ gives rise to an exact sequence $1\to Ker\nu \to \mathbf F_p^*\to \mathbf F_p^* \to Coker \nu \to 1$. Since the groups on play are all finite, the kernel and cokernel have the same order. But $\mathbf F_p^*$ is cyclic, hence the order of $Ker\nu$ is $d:=gcd(n, p-1)$. In conclusion, if $p$ is odd and $n=mp^k$ with $p$ not dividing $m$, $Hom({\mathbf Z_p}^*, \mathbf Z /n) \cong (\mathbf Z/p^k) \times (\mathbf Z/d)$.