Let G be $\mathbb Z_p\times\dots\times \mathbb Z_p$ . Find A(G). Let G be $\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}$. Find $A(G)$. 
I know that $A(G)\cong GL_n(\mathbb Z_p)$. 
I prove it by taking $\varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $\theta$ from $A(G)$ and I show that the composition
of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible. 
I don't know if this is good way or not, I know it's good for $\mathbb Z_p\times\mathbb Z_p$ but I don't know if it's enough for $\mathbb Z_p\times \mathbb Z_p\times\dots\times\mathbb Z_p$ (n times). 
If you have other wat please help me, there is a hint to think about vector space $(\mathbb Z_p)^n$ but I don't know it could help. 
 A: Supposing that $\mathbb{Z}_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(\mathbb{Z}_p)^n$ is also a continuous $\mathbb{Z}_p$-module automorphism (as we can see the $\mathbb{Z}_p$ action as limits of $\mathbb{Z}$ actions). So we have:
$$A((\mathbb{Z}_p)^n) = A_{\mathbb{Z}_p}((\mathbb{Z}_p)^n)$$
But now, $(\mathbb{Z}_p)^n$ is a free $\mathbb{Z}_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $\mathbb{Z}_p$ is not a field). Moreover, fixing a $\mathbb{Z}_p$-basis of $(\mathbb{Z}_p)^n$ (for example, the canonical one), every $\mathbb{Z}_p$-homomorphism of $(\mathbb{Z}_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $\mathbb{Z}_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_{\mathbb{Z}_p}((\mathbb{Z}_p)^n) \simeq \text{GL}_n(\mathbb{Z}_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.

If $\mathbb{Z}_p$ means the integers modulo $p$, notice that $(\mathbb{Z}_p)^n$ is a vector space over $\mathbb{Z}_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $\mathbb{Z}_p$ action as a quotient of a $\mathbb{Z}$ action). So $A((\mathbb{Z}_p)^n)$ is equal to the space of all invertible linear transformations os $(\mathbb{Z}_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
