Suppose a finite--dimensional Lie group $G$ is given. Does there exist a manifold $M$ and a Riemannian metric $g$, such that $G$ is the **full** isometry group of $(M,g)$?

For example if I try to do this for a connected $G$, then I often get a bigger group as the full isometry group, which includes e.g. the orientation reversing isometries. (Maybe one has to take a non--orientable space for that?)

Even if I try to realize $\mathbb R$ as a full isometry group, I fail. (One could take the full isometry group of $\mathbb R$ with the standard metric, which is given by $\mathbb R \ltimes \mathbb Z_2$ and divide out the $\mathbb Z_2$ action. But this leads to a fixpoint and the quotient is therefore not a manifold any more.)

There is an article of J. de Groot which proves that every abstract group can be realized as an isometry group of some metric space, but it is not clear to me, if this is true in the category of Lie groups and Riemannian manifolds.

Greetings James