Smooth embedding of space forms in the Euclidean space

I was wondering which $$S^n/\Gamma$$ can be smoothly embedded into $$\mathbb R^{n+1}$$, where $$\Gamma \subset O(n+1)$$ is a finite subgroup. To my knowledge, the case $$n \le 3$$ is known. It has been proved that when $$n=3$$, this is possible only if $$\Gamma=1$$ or $$\Gamma=Q_8$$.

If $$n \ge 4$$, is there any nontrivial group $$\Gamma$$ so that the embedding is possible?

You can certainly show that many such space forms do not embed. The simplest version would be that a lens space ($$L =S^n/\mathbb{Z}_k$$; here $$n$$ should be odd) whose fundamental group has order $$k=p^r$$ for $$p$$ a prime does not embed. To see this, replace $$\mathbb{R}^{n+1}$$ with $$S^{n+1}$$ that would be split into two pieces, say $$A$$ and $$B$$, by $$L$$. Then the homology of $$L$$ is the direct sum of the homology groups of $$A$$ and $$B$$ by an easy Mayer-Vietoris argument. In particular, either $$A$$ or $$B$$ would be a null-bordism over $$\mathbb{Z}_k$$ for $$L$$, which represents the generator of $$H_n(B\mathbb{Z}_k) = \mathbb{Z}_k$$. The observation about the homology goes back quite a while; cf. Hantzsche, W.; Einlagerung von Mannigfaltigkeiten in euklidische Räume. Math. Z. 43 (1938), no. 1, 38–58.
I imagine that this would work for composite $$k$$ with a slightly more elaborate argument. There are also lots of other obstructions; unlike in the classical case $$S^n/\Gamma$$ might not be stably parallelizable, so there is no codimension-one embedding. In fact, it's hard to find stably parallelizable lens spaces; cf. Ewing, John; Moolgavkar, Suresh; Smith, Larry; Stong, R. E. Stable parallelizability of lens spaces. J. Pure Appl. Algebra 10 (1977/78), no. 2, 177–191.