The answer is yes for finite groups -- you can even ensure your space is a hyperbolic surface, or hyperbolic 3-manifold.  The result for 3-manifolds is Sadayoshi Kojima's.  For hyperbolic surfaces I forget who its due to, but I think the reference is in Kojima's paper. 

For compact Lie groups I think you can just make your space some variant of the Lie group itself (with a left-invariant metric).  The idea would be to take the Lie group $G$ with its left-invariant metric.  If that has a bigger isometry group than $G$ itself, take the product of $G$ with a ball (with some metric), and perturb the metric on $G \times B$, $G$-equivariantly.  Some generic perturbation of the metric should kill all isometries other than the ones coming from the $G$-action. 

Andre raises the question of whether or not this construction could be performed for $S^1 \equiv SO_2$.   The issue being that with its left-invariant metric the isometry group is $O_2$.  I don't see any reason why it shouldn't work.  For example, take $S^1 \times D^2$.  We don't put quite the product metric on $S^1 \times D^2$ but we do put a metric on it that's locally a product.  The idea is to put a metric on the fibres which is a disc with only one isometry, which is a rotation by $\pi$ about the centre -- so there is a preferred axis in the disc, and the isometry of the disc flips the orientation of the axis.  On $S^1 \times D^2$ you put the metric where the disc's axis does a rotation by $\pi$ as one goes about the base $S^1$.