Direct limit of automorphism group of cyclic groups of order $p^r$. Let ${\rm Aut}((\mathbf{Z}/p^r))$, denote the automorphism group of the cyclic group of order $p^r$. Then $\{ \mathbf{Z}/p^r, i^r\}_{r \in \mathbf{N}}$, with inclusions $i^r : \mathbf{Z}/p^r \to \mathbf{Z}/p^{r+1}$ forms a direct system. Then can we identify (up to isomorphism) the automorphism groups ${\rm Aut}(\mathbf{Z}_{p}) = {\rm colim}_{r \in \mathbf{N}} {\rm Aut}(\mathbf{Z}/p^r)$, where $\mathbf{Z}_p$ denotes the group of $p$-adic integers?
 A: Yes, it works and also for its powers, and endomorphisms: if $M,M'$ are finitely generated $\mathbf{Z}_p$-modules, then writing $M_n=M/p^nM$, we have, denoting by Hom group homomorphisms
$$\mathrm{Hom}(M,M')=\mathrm{Hom}(M,\underleftarrow{\lim}M'_n)=\underleftarrow{\lim}\mathrm{Hom}(M,M'_n)=\underleftarrow{\lim}\mathrm{Hom}(M_n,M'_n).$$
In particular, we have an injective natural map $\underleftarrow{\lim}\mathrm{Aut}(M_n)\to \mathrm{Aut}(M)$, and in the reverse direction a natural monoid homomorphism map $\mathrm{Aut}(M)\to\underleftarrow{\lim}\mathrm{End}(M_n)$; since the composite map into $\mathrm{End}(M_n)$ maps into the subgroup of invertible elements, the previous map maps into $\underleftarrow{\lim}\mathrm{Aut}(M_n)$. So we have a natural isomorphism $\mathrm{Aut}(M)\to\underleftarrow{\lim}\mathrm{Aut}(M_n)$.

Let $M$ be an abelian group such that $M=\bigcup M_{(n)}$, where $M_{(n)}$ is the kernel of multiplication by $p^n$. Then the structural ring homomorphism $\mathbf{Z}\to\mathrm{End}(M)$ has a unique extension to $\mathbf{Z}_p$.
Suppose moreover that $M_{(n)}$ is cyclic of order $p^n$ for all $n$. Then the resulting continuous homomorphism $\mathbf{Z}_p\to\mathrm{End}(M)$ is a topological isomorphism. It is continuous and injective on $\mathbf{Z}$; since the only closed subgroups of $\mathbf{Z}_p$ are the $p^n\mathbf{Z}_p$, the kernel is zero. For surjectivity, by compactness, it is enough to check that it has a dense image ($\mathrm{End}(M)$ is endowed with the topology of pointwise convergence). And indeed every finite subset is contained in $M_{(n)}$ for some $n$, and every endomorphism coincides with some integral multiplication on $M_{(n)}$. So the image of $\mathbf{Z}$ in $\mathrm{End}(M)$ is dense.
In particular, the canonical continuous map $\mathbf{Z}_p^\times\to\mathrm{Aut}(M)$ is a topological isomorphism in this case.
