I've been struggling with this one for a couple of hours.

The group (R,+) has (R-{0},*) as its automorphism group. The positive reals are a subgroup of index 2 isomorphic to (R,+) via the mapping x->log(x). The same mapping shows that there is a subgroup (R-{0},*) isomorphic to (Q,+). This shows that it is not completely implausible that a group G might have Aut(G) isomorphic to (Q,+)

On the other hand there are plenty of things which make it look like it is probably False.

Clearly G must be infinite since otherwise Aut(G) is finite.

The mapping g->g^-1 is an automorphism of order 2 if G is abelian unless every element of G has order 2. Since (Q,+) is torsion free then either G is abelian and every element has order 2, or G is non-abelian.

In the former case G cannot be finitely generated so we would expect Aut(G) to be isomorphic to a subgroup of the invertible linear transformations of an infinite dimensional vector space over F2. It is not easy to see how such a subgroup could possibly be isomorphic to (Q,+) : none of the transformations that fix all but finitely many generators can be in the subgroup, since these all have finite order.

In the latter case (G not abelian) there are distinct elements a,A such that g->A*g*a is a non-trivial automorphism (so a,A are inverses of each other). All such elements must be torsion-free since otherwise there would be an automorphism with finite order. Now the requirement that Aut(G) is isomorphic to (Q,+) means that for all (a,b,g in G) we have B*A*g*a*b=A*B*g*b*a (again using case change for inverse pairs of elements).

So G is probably not finitely presented even if it were finitely generated. Also with so many torsion-free elements one would expect to be able to find an automorphism that permuted finitely many generators and thus had finite order.