Let $g$ be a metric of arbitrary signature on $\Bbb R^n$. If the Riemann tensor of $g$ vanishes identically, is there a diffeomorphism $f:\Bbb R^n\to\Bbb R^n$ with $g=f^*e$, where $e$ is the standard metric with the same signature as $g$?
In the Riemannian case, assuming $(\Bbb R^n,g)$ is complete, this is well known from the classification of space forms. I was unable to show that $(\Bbb R^n,g)$ is always complete in the Riemannian case, and I don't know if it's true or not. I also know that this is always true in a local sense, but I am looking for a global isometry.
Pick any diffeomorphism $\phi$ from $\Bbb R^n$ to a proper (necessarily simply connected) open subset $U \subset \Bbb R^n$. (For concreteness, we could take $$U := (-1, 1)^n, \qquad \phi : (x_1, \ldots, x_n) \mapsto (\tanh x_1, \ldots, \tanh x_n)\textrm{.)}$$ Then, the pullback $\phi^* (e \vert_U)$ is a flat but incomplete metric on $\Bbb R^n$, and in particular it is not globally isometric to $(\Bbb R^n, e)$.