Question

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?

Answer 1

After all these comments, a possible answer to your question goes as follows.
If $n$ is even, then the only group that can act freely and isometrically on $S^n$ is $\mathbb{Z}/2$. One way to see this is as in Max's answer above, i.e. by looking at the behavior of eingenvalues of orthogonal matrices. Here is another reason: an action of $G$ on $S^n$ means that there is a group homomorphism $G\rightarrow\mbox{Homeo}(S^n)$. But any homeomorphsim has degree $\pm 1$. So we get a homomorphism $G\rightarrow\mathbb{Z}/2$. But the action has no fixed points, then the degree of the image of every non trivial element in $G$ is $(-1)^{n+1}=-1$, therefore the map has a trivial kernel and therefore it is an isomorphism.
As macbeth already pointed out, when $n$ is odd, the problem is not so easy. In that case, we consider $S^n$ as the universal cover of a complete Riemannian manifold $M$ with constant sectional curvature. A simple argument using lifting properties of covering spaces shows that there is an isometry between $M$ and $S^n/G$ (whenever the action is free and properly discontinuous). So the problem of finding the subgrups with that particular action on the sphere is the same as the classification of complete Riemannian manifolds with constant sectional curvature (=1). That is done in Wolf's book Spaces of Constant Curvature.

Answer 2

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 non-trivial 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 non-trivial 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 $n$.