I proved this in dimensions $n = 1, 2, 3$ from the direct classification of subgroups. For $n \geq 4$ I can't understand anything. Also, I can't find any information about this issue. Is this a solved problem or an open problem?
Any discrete $G$ can be realized: take a sufficiently small affinely independent finite set of points far from $0$ and act on it with $G$. It is easy to prove that the symmetry group of the resulting compact set coincides with $G$. If $G$ is not discrete, then I do not understand whether this remains true.
As requested in the comments, proof for n = 3: Since the Lie algebra of (the 3-dimensional Lie group) $SO(3) \cong S^3 /\{\pm\}$ is $\mathbb{R}^3$ with vector product (which has no 2-dimensional subalgebras), then $SO(3) $ has no closed subgroups of dimension 2. Subgroups of dimension 0 are discrete, a subgroup of dimension 3 is the whole group. Due to the bijection between the immersion connected subgroups of the Lie group and the subalgebras of its algebras, we have that the connected component of any one-dimensional subgroup of $SO(3)$ is the group of rotations around a line (and there are no other immersion subgroups, because the groups of rotations around the line have already occupied all the one-dimensional subalgebras). It is easy to see that then any one-dimensional subgroup is either a group of rotations of a cone or a group of rotations of a cylinder (in other words, any overgroup of a subgroup of rotations about a line containing an element that does not preserve this line coincides with $SO(3)$).
$SO(4) \cong (S^3 \times S^3) /\{\pm 1\}$, in particular, its Lie algebra is the direct product of $\mathbb{R}^3 \times \mathbb{R}^3 $ (with componentwise vector product). Thus subgroups have dimension $0, 1, 2, 3, 4, 6$. I can't figure out how to analyze subgroups of dimensions $1, 2, 3, 4$ in this case.