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 $S^1$.

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 $S^1 \times D^2$. Put a metric on this space which is locally a product, but where the fibre is a disc whose isometry group is $\mathbb Z_3$, orientation-preserving isometries of a triangle. We make the holonomy of the bundle around the base circle the generator of this isometry group. So there can be no isometry of this bundle that reverses the direction of the base space, since that would mean the holonomy is equal to its inverse, which in $\mathbb Z_3$ can not happen. So the only symmetries of this bundle act as orientation-preserving isometries of the base, and orientation-preserving on the fibre, and this group is $SO_2$.

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.

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 $S^1$.