Yes, you are one-hundred percent correct. Namely, $\text{Aut}(\mathbb{Z}_p^n)\cong\text{GL}_n(\mathbb{Z}_p)$. Now, for the second part let's let $G=\text{GL}_n(\mathbb{Z}_p)$, $X=\mathbb{Z}_p^n$, let $G$ act on $X$ in the usual way. Take some $v\in X$ not equal to zero. Begin then by noting that $Gv=X-\{0\}$. Let's see if we can find what the stabilizer of $v$ is. Well, imagine taking a matrix in $\text{stab}(v)$ with respect to some basis $\{v,x_1,\cdots,x_{n-1}\}$ containing $v$. Now, we know that if we take the first column of this matrix with respect to $v$ we have $(1,0,\cdots,0)$ since $v$ maps to itself. Now, the only requirement on the second column is that it doesn't break invertibility, which is to say that it is linearly independent from the first column. Now, this is equivalent to saying that $x_1\notin\text{span}_{\mathbb{Z}_p})(v)$. But, since $|\text{span}_{\mathbb{Z}_p}(v)|=p$ there are $p^n-p$ such choices for $x_1$. Now, similarly, the only condition on choosing $x_2$ is that we require invertibility, which means choosing $x_2$ to be linearly independent from $\{v,x_1\}$ which means choosing $x_2$ not in $\text{span}_{\mathbb{Z}_p}(v,x_1)$. Since $|\text{span}_{\mathbb{Z}_p}\{v,x_1\}|=p^n-p^2$ we see that there are $p^n-p^2$ choices for $x_2$. Continuing in this fashion we find that the number of ways to pick the $2,\cdots,n$ columns of a matrix in $\text{stab}(v)$ are $(p^n-p)\cdots(p^n-p^{n-1})$ and so $|\text{stab}(v)|=(p^n-p)\cdots(p^n-p^{n-1})$. Now, the orbit stabilizer theorem tells us
$|G|=|Gv||\text{stab}{v}|=(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})$