If a metric tensor is not conformally equivalent to the flat metric If on a manifold $M$ we have two metrics $g_{ab}$ and $g'_{ab},$ which are not conformally equivalent, and we say that $(M,g_{ab})$ is a flat manifold, does it follow that $(M,g'_{ab})$ is not flat? Or are there cases where there is still a mapping of $(M,g'_{ab})$ into flat space?
Does this change if we consider $g'_{ab}$ on a different manifold $M'$, such that $g_{ab}$ and $g'_{ab}$ are not conformally equivalent?

Certainly not.  For instance, take $M=\mathbb{R}^2$ and let $f:M\to M$ be any diffeomorphism which is not conformal.  Let $g_{ab}$ be the standard metric, and let $g'_{ab}=f^*g_{ab}$.  Then $g_{ab}$ and $g'_{ab}$ are both flat, but they are not conformally equivalent because $f$ is not conformal.
Less obviously, you can get a counterexample where $(M,g_{ab})$ and $(M,g'_{ab})$ are not even conformally equivalent by any map (i.e., there is no diffeomorphism $f:M\to M$ such that $g'_{ab}$ is conformally equivalent to $f^*g_{ab}$).  You can get such an example by looking at $2$-dimensional tori, for instance.  Every lattice $\Lambda\subset\mathbb{R}^2$ gives a torus $\mathbb{R}^2/\Lambda$ which inherits a flat metric from $\mathbb{R}^2$.  These tori are all diffeomorphic (so they can be considered as different metrics on the same underlying manifold), but different lattices usually give conformally inequivalent tori.
