Diagonal group action on a product manifold: when is the quotient a product manifold? Suppose $M$ and $N$ are manifolds, each acted on freely by a group $G$. If it helps, I'm happy to assume $M$ and $N$ are compact and $G$ is a finite group (or a compact Lie group). Consider the "diagonal" action of $G$ on the product manifold $M \times N$, defined by $g \cdot (m, n) := (g \cdot m, g \cdot n)$. 

What can be said about the topology of the quotient?
In particular, when is $(M \times N)/G$ homeomorphic to a product of manifolds, and when is it homeomorphic to either of the products $M/G \times N$ and/or $M \times N/G$?
I suppose the Künneth theorem gives necessary conditions for the quotient to be a product. Are there
  other necessary and/or sufficient conditions?
Can more be said if $M = N$ and the action of $G$ is the same on both?

If not much can be said in general, I am especially interested in spheres and lens spaces. For simplicity, let's say $M$ and $N$ are spheres and $G = \mathbb{Z}_2$ consists of the antipodal map and the identity (so then $M/G$ and $N/G$ are each a real projective space).
For example, suppose $M = S^3$ and $N = S^5$. I think the Künneth theorem shows that $\mathbb{RP}^3 \times S^5$ and $S^3 \times \mathbb{RP}^5$ have different integral homology groups, so they are not homeomorphic. So in this case, $(M \times N)/G$ cannot be homeomorphic both to $M/G \times N$ and to $M \times N/G$.
We could also consider spheres of the same dimension. For example, if $M = N = S^1$, then I think (from drawing a picture) that the quotient is homeomorphic to the torus $S^1 \times S^1$. (But maybe this is just luck due to the low dimension. Note $\mathbb{RP}^1 = S^1/\mathbb{Z}_2$ is just $S^1$ again.) If $M = N = S^3$, is the quotient homeomorphic to $\mathbb{RP}^3 \times S^3$? What if only one of the factors is $S^1$?
 A: In general, all you can say is that $(M\times N)/G$ is the total space of of two bundles simultaneously:
$N\rightarrow (M\times N)/G \rightarrow M/G$ and $M\rightarrow (M\times N)/G\rightarrow N/G$.  This, of course, puts strict constraints on the cohomology ring and homotopy groups of $(M\times N)/G$, but does not neccesarily force $(M\times N)/G$ to be a product, even up to homotopy.
The case of spheres with antipodal actions is interesting.  If both spheres are even dimensional, then $(M\times N)/G$ is orientable, while $M/G$ and $N/G$ are not, so there is no way $(M\times N)/G$ is homotopy equivalent to either $M\times (N/G)$ or to $(M/G)\times N$.
Further, $(S^{2k}\times S^1)/G$ is not homotopy equivalent to either a) $\mathbb{R}P^{2k}\times S^1$ or to b) $S^{2k}\times \mathbb{R}P^1 = S^{2k}\times S^1$.  To see this, first note that $(S^{2k}\times S^1)/X$ is non-orientable, ruling out b).
Also, since its an $S^{2k}$ bundle over $S^1$, the long exact sequence in homotopy groups shows $\pi_1((S^{2k}\times S^1)/G)$ is isomorphic to $\mathbb{Z}$, so it can't be homotopy equivalent to the possibility a).
On the other hand, $(S^3\times S^3)/G$ and $(S^{2k+1}\times S^1)/G$ are diffeomorphic to $S^3\times \mathbb{R}P^3$ and to $S^{2k+1}\times S^1$, respectively.  For the $(S^3\times S^3)$ we note that $(S^3\times S^3)/G$ is the Lie group $SO(4)$.  Then one has the bundle $SO(3)\rightarrow SO(4)\rightarrow S^3$.  As all bundles with finite dimensional structure group are trivial over $S^3$, this bundle is trivial.  Finally, since $SO(3)$ is diffeomorphic to $\mathbb{R}P^3$, this gives the desired diffeomorphism.
For $(S^{2k+1}\times S^1)/G$, we first note this space is orientable.  Further, we get a bundle $S^{2k+1}\rightarrow (S^{2k+1}\times S^1)/G\rightarrow S^1/G = S^1$.  Since the total space is orientable, from the classification of bundles over $S^1$, this bundle is trivial (at least, using homeomorphisms to trivialize instead of diffeomorphisms).
I don't know what happens with other products of spheres.
