Finite abelian group whose automorphism group is some general linear group
Theorem: Let $p$ be a prime and $G$ be a $p$-group. If ${\rm Aut}(G)$ is isomorphic to $GL(n, p)$ for some positive integer $n$, then $G$ is an elementary abelian group of order $p^n$.
Reference: General Linear Groups as Automorphism Groups
Let me write down a couple of thoughts (not yet a proper answer) and I'll get back to this problem a little later, as I have to leave now.
First, the automorphism group of a $p$-group is never trivial unless the group has order $2$.
Second, since an abelian group is the direct product of its Sylow subgroups, and since automorphism groups behave well with respect to direct products of groups of coprime orders, it follows that the automorphism group of an abelian group is the direct product of the automorphism groups of its Sylow subgroups.
Third, no general linear group is directly decomposable, i.e. you can't write a general linear group as a direct product of proper subgroups.
So if there is a minimal counterexample, then it must be a $p$-group.
At this point, you need to know how to find the order (but perhaps not necessarily the structure) of the automorphism group of an abelian $p$-group. This is given by Theorem 4.1 of this paper.
Assuming that $G = C_{p^{e_1}} \times \ldots \times C_{p^{e_r}}$, where $e_1 \leq \ldots \leq e_r$, we have
$$\lvert \operatorname{Aut}(G) \rvert = \prod_{k=1}^r(p^{d_k}-p^{k-1})\prod_{j=1}^r p^{e_j(r-d_j)}\prod_{i=1}^r p^{(e_i-1)(r-c_i+1)},$$ where $d_k = \operatorname{max}\{s:e_s=e_k\}$ and $c_k = \operatorname{min}\{s:e_s=e_k\}$, for $1 \leq k \leq r$.
We are also assuming that $\lvert \operatorname{Aut}(G) \rvert = \lvert \operatorname{GL}_n(2) \rvert$ for some positive integer $n>1$. Thus the problem essentially reduces to showing that if $$\lvert \operatorname{Aut}(G) \rvert = \prod_{k=1}^r(p^{d_k}-p^{k-1})\prod_{j=1}^r p^{e_j(r-d_j)}\prod_{i=1}^r p^{(e_i-1)(r-c_i+1)} = \lvert \operatorname{GL}_n(2) \rvert = 2^{\frac{n(n-1)}{2}}\prod_{i=1}^n (2^i-1),$$ then $p =2$, $n=r$ and $e_1 = \ldots = e_r=1$.
Let $$p^A =\prod_{j=1}^r p^{e_j(r-d_j)}\prod_{i=1}^r p^{(e_i-1)(r-c_i+1)}.$$ Then $$p^Ap^{\frac{r(r-1)}{2}}\prod_{k=1}^r(p^{d_k-(k-1)}-1) = 2^{\frac{n(n-1)}{2}}\prod_{i=1}^n (2^i-1).$$
By using the result cited by Dietrich Burde, we can assume that $p>2$. Then $2^{\frac{n(n-1)}{2}}$ must divide the quantity $\prod_{k=1}^r(p^{d_k-(k-1)}-1)$. I'll pick this up tomorrow.
I am not sure that this approach leads anywhere. Meanwhile, I have mentioned this problem to Marty Isaacs and he destroyed it effortlessly. Here's his solution:
Suppose that $G$ is finite abelian but not an elementary abelian $2$-group and such that $\operatorname{Aut}(G) \cong \operatorname{GL}_n(2)$, $n>1$. The map $g \to g^{-1}$ is an automorphism of $G$, has order $2$ (since $G$ is not an elementary abelian $2$-group) and lies in the centre of $\operatorname{Aut}(G)$ (check this). But $\operatorname{GL}_n(2)$ has trivial centre, contradiction. Thus $G$ is an elementary abelian $2$-group and since the sequence $\lvert \operatorname{GL}_n(2) \rvert$ is strictly increasing in $n$, we get $|G| = 2^n$ thus $G \cong C_2^n$.