Sorry for the poor notations, but I'd greatly appreciate any comments/suggestions for my proof written below.
We know for a fact that $\Bbb Z_{p}^d$ is a vector space of dimension $d$ over the field $GF(p)$, and $$GL(d,p)=\{\text{invertible }d\times d\text{ matrices over the field } GF(p)\}$$
On the other hand $$Aut(\Bbb Z_{p}^d) = \{\text{bijective homomorphisms from }\Bbb Z_p^d\text{ to itself}\} = \{\text{linear transformations }T :\Bbb Z_p^d \to \Bbb Z_p^d \text{ such that $T$ is invertible}\}$$
Therefore $Aut(\Bbb Z_{p}^d)$ is isomorphic to $GL(d,p)$.
How can I construct a more rigorous proof for this lemma?
There are two groups here.
There is $G_1$, say: the automorphisms of the abelian group $(\mathbb{Z}_p)^d$.
There is $G_2$, say: the automorphisms of the $\mathbb{Z}_p$-vector space $(\mathbb{Z}_p)^d$, isomorphic of course to the matrix group $\text{GL}(d,p)$ via a choice of basis.
Now obviously every element of $G_2$ is an element of $G_1$; just forget the scalar multiplication, it still preserves the group addition.
But the converse is true. Given an element $\alpha$ of $G_1$, it preserves all the additive group structure, but it also preserves the scalar multiplication, since $$ \alpha(r\cdot v)=\alpha(v+\dots+v)=\alpha(v)+\dots+\alpha(v)=r\cdot\alpha(v) $$ where the $\dots$ mean a sum with $r$ terms.
So $G_1=G_2$.