Is there some sort of classification of finite groups $G$ such that for at least one $n$ the group $G$ admit a free isometric action on the standard sphere $S^n $of curvature 1? Are there some simple criteria that permit to check (in some particular cases) if a given group has such an action (for at least on $n$) or not?

After all these comments, a possible answer to your question goes as follows. 


EDIT 2: My original and the following revised answer were nonsense; I'll just leave this edited, and very incomplete "answer" dealing with even $n$, just in case it helps somebody avoid to repeat my mistakes :(. The isometry group of the standard sphere $S^n$ is the orthogonal group $G:=O_{n+1}(\mathbb{R})$. Each element of $G$ is either a rotation, a reflection or the product of a rotation and a reflection. Every reflection fixes a hyperplane, hence also a point on the sphere. Thus a group $H$ acting freely and isometrically cannot contain reflections. Now if $n$ is even (and hence $n+1$ odd), then every rotation of $G$ fixes a point. (Its eigenvalues have absolute value 1, and it must have at least one real eigenvalue, hence 1 or 1. But the complex eigenvalues come in pairs $\lambda,\overline\lambda$. Since rotations have determinant 1, we conclude that the eigenvalue $1$ must occur an even time. Hence there is an eigenvector with eigenvalue 1.) This leaves products of a reflection and a rotation (such as the antipodal map $x\mapsto x$). One can then show that at most one such nontrivial element can be contained in $H$ (as otherwise, we would get reflections or rotations inside $H$), and in fact, only the antipodal map can occur as nontrivial element of $H$. Thus only the trivial group and the cyclic group of order 2 can act freely and isometrically on $S^n$ for even $n$. On the other hand, for odd $n$, more possibilities arise, e.g. any cyclic group admits a free isometric action on $S^n$ for odd $n$. 

