Let $M,N$ be smooth **Riemannian** manifolds with **boundary** (In particular, we assume the boundaries are smooth).

Suppose we have a map $\phi:M \to N$ which satisfies the following properties:

$$(1) \, \, \phi:M \to N \, \, \text{is a bijection}$$

$$ (2) \, \, \phi(\operatorname{int}M)=\operatorname{int}N,\phi(\partial M)=\partial N $$

$$ (3) \, \, \phi:M \to N \, \,\text{is a metric isometry}$$

By the Myers-steenrod theorem, applied to $\phi|_{\operatorname{int}M} :\operatorname{int} M \to \operatorname{int}N $, $\phi$ is a diffeomorphism between $\operatorname{int} M , \operatorname{int}N$.


 **Question: Is $\phi$ necessarily smooth as a map $M \to N$?**
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When looking at the proof of Myers-steenrod theorem [here][1], the problem seems to be that **initial conditions do not determine a unique geodesic, if the starting point is on the boundary.** 

The basic idea of the proof is to express $\phi$ in exponential coordinates, then show that this representation is linear, hence smooth. However, constructing this representation relies on the uniqueness of geodesics. 

I suspect there might be a counter example where singularity can occur at the boundary, but I could not fine one.

[1]:https://amathew.wordpress.com/2009/11/17/isometries-of-riemannian-manifolds-and-a-theorem-of-myers-steenrod/