Based on the following proof my attempt is to prove that:
If $G$ is $p$-elementary abelian then ${\rm Aut}(G)$ acts transitively on $G\setminus\{1\}$.
I'm stuck on this. But I tried a few things, for example, I know that every $p$-elementary abelian group has a $\mathbb{Z}_p$-vector space structure and $G \simeq \mathbb{Z}_p^n$ for some $n$. So viewing $G$ as a vector space we have that the automorphisms of $G$ are precisely the linear transformations $T:G\to G$ with $\ker(T) = 0$. Then $$ {\rm Aut}(G) = \{T:G\to G \mid ker(T) = 0\} \simeq GL_n(\mathbb{Z}_p). $$ So I need to prove that the general linear group $GL_n(\mathbb{Z}_p)$ acts transitively on $\mathbb{Z}_p^n \setminus \{0\}$. How do I do it?
You're exactly right: this is equivalent to showing that $\mathrm{GL}_n(\mathbb{Z}_p)$ acts transitively on $\mathbb{Z}_p^n \setminus \{0\}$. There's nothing special about $\mathbb{Z}_p$ here, actually, this is true over any field!
Claim Let $k$ be a field and let $n$ be a natural number. Then $\mathrm{GL}_n(k)$ acts transitively on $k^n \setminus \{0\}$.
To prove this, you need to take arbitrary elements $v, w \in k^n \setminus \{0\}$ and then construct an element $T \in \mathrm{GL}_n(k)$ such that $T(v) = w$. Remember: a linear transformation is uniquely determined by its action on a basis. Moreover, a linear transformation is invertible iff it sends bases to bases.
So, you want to pick bases $\beta$ and $\beta'$ for $k^n$, and then define $T$ by mapping $\beta$ (bijectively) to $\beta'$. If you make convenient choices for $\beta$ and $\beta'$, it will be easy to ensure that $v$ is sent to $w$!