Let $ (M,g) $ be a flat and Riemannian homogeneous (isometry group acts transitively) manifold. Then is $ M $ just diffeomorphic to a product of circles and lines? This is true in dimension 2 (and 1).
Specifically, is every flat compact Riemannian homogeneous manifold a flat torus? I know by theorem of Bieberbach that every flat compact manifold is covered by a flat torus. So the idea would be that any nontrivial covering breaks the symmetry.
A Riemannian homogeneous manifold is diffeomorphic to the product of a Euclidean space with a compact Riemannian homogeneous manifold. See
https://mathoverflow.net/questions/410334/noncompact-riemannian-homogeneous-is-trivial-vector-bundle-over-compact-homogene
So it is enough to show this for the compact case.
Let $ M $ be a flat compact Riemannian homogeneous manifold. Then the fundamental group of $ M $ is a torsion free crystallographic group (also known as a Bieberbach group). Since $ M $ is compact and Riemannian homogeneous then by
Transitive action by compact Lie group implies almost abelian fundamental group
the commutator subgroup of the fundamental group is finite. But the fundamental group is torsion free so every finite subgroup is trivial. Thus the commutator subgroup must be trivial. In other words $ \pi_1(M) $ is abelian. So it must be $ \mathbb{Z}^n $. So $ M $ must be isometric to a flat torus.