Let $M,N$ be smooth, connected, oriented $d$-dimensional Riemannian manifolds (perhaps with boundary).
Suppose there exist a map $f:M \to N$ which is differentiable almost everywhere on $M$, and its differential is an orientation-preserving isometry almost everywhere.
Is it true that $M$ can be isometrically immersed in $N$? (i.e does there exist a smooth isometric immersion $M \to N$?)
If we omit the requirement of orientation-preserving, then this is false:
Gromov showed that for any metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an arcwise isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$ (which has to be an isometry a.e).
If $g$ is non-flat, then no smooth isometric immersion exist, of course.
Note: In that case, Gromov's arcwise isometry $f$ cannot preserve orientation: It is $1$-Lipschitz, and hence in $W^{1,\infty}(M,N)$. My colleagues and I showed here (section 3) that any weakly differentiable map $f \in W^{1,\infty}(M,N)$ satisfying $df \in SO(n)$ a.e is weakly harmonic, hence smooth. (This shows $f$ cannot be orientation-preserving or orientation-reversing on any open subset of the domain. It must "switch" rotations infinitely often).
Edit: It turns out $M$ is not necessarily isometrically immersible in $N$. However, in the example given $M,N$ were both flat.
This still leaves open the question whether examples of such maps exist for pairs of maniolds which have different curvatures.
Specifically, is there an example for a non-flat $M$, and a flat $N$, such that there exist an a.e orientation-preserving isometric immersion $M \to N$?
The point is to see how much flexibility this notion allows. Gromov's maps show that curvature differences between manifolds do not obstruct existence of a.e isometries.
Do such differences obstruct a.e-orientation-preserving isometries?
Note that in general an "almost everywhere isometry" does not need to be smooth.
An example is $M=[0,1],N=[0,2],f(x)=c(x)+x$ where $c$ is the Cantor function. In this case, $f'=1$ a.e, and $M$ can be isometrically immersed in $N$, via the natural inclusion map.
To refute the claim, one could try to use $f^{-1}:[0,2] \to [0,1]$, but it is not differentiable a.e on $[0,2]$;
$f ^{-1}$ is differentiable on $f(O)$, when $O$ is the complement of the Cantor set (since $O$ is the set where $f$ is differentiable). By the definition of $f$, it is not hard to see that $m(f(O))=m(O)=1$, hence is not of full measure in $[0,2]$.
This answer is based on a comment by Moishe Cohen.
Let $M,N$be circles with radiuses $1,2$ respectively. We define $g:M \to N$ by
$$g(e^{2\pi it})=2e^{\pi if(t)}$$
where $f(x)=x+c(x)$ is the example function given in the question.
$g$ is differentiable at $\{e^{2\pi it} \, | \,t \in [0,1]\setminus C \}$, which is a set of full measure on the unit circle $M$.
Let $ \alpha(s)=e^{2\pi i(t+s)},\alpha'(s)=2\pi i e^{2\pi i(t+s)},\alpha'(0)=2\pi i e^{2\pi it}$: Then
$$dg_{(e^{2\pi it})}(2\pi i e^{2\pi it})=\frac{d}{ds}|_{s=0}g(\alpha(s))=\frac{d}{ds}|_{s=0} 2e^{\pi if(t+s)}=2e^{\pi if(t)}f'(t)=2e^{\pi if(t)}$$
for every $t \in [0,1]\setminus C $.
So $dg_{(e^{2\pi it})}$ maps a vector with length $2\pi$ to a vector of the same length, and since the tangent spaces are one-dimensional this implies it's an isometry a.e.
It is also easy to see it's differential is orientaion-preserving whenever it is defined.
Thus, $g$ satisfies the requirements, but $M$ is not isometrically immersible in $N$.