Let $M$ be a compact, connected, oriented, smooth Riemannian manifold with non-empty boundary. Let $f:M \to M$ be a smooth orientation preserving isometric immersion.
Is it true that $f(\partial M) \subseteq \partial M$?
Edit: In this paper, it is proved that any "local isometry" from a compact, connected metric space into itself is a homeomoprhism, where by a "local isometry", they mean "locally preserving distance".
Thus, if every isometric immersion is locally preserving distance, then our $f$ will be a homeomorphism (hence will map boundary to boundary).
Is it true that isometric immersion $\Rightarrow$ locally preserving distance?
Motivation:
A positive answer to this question would imply that every smooth orientation preserving isometric imersion $M \to M$ is a Riemannian isometry (see details below**), Moreover, if two manifolds can be isometrically immersed in each other, then they are isometric (see remark 1 in the "updates and remarks" of this question).
In particular, to refute the conjecture "$f(\partial M) \subseteq \partial M$" it is enough to find such an immersion which is not surjective, or at least not an isometry.
Remarks:
(1) Compactness is essential. Look at $M=[0,\infty) \,,\,f(x)=x+1$.
(2) The claim clearly holds in dimension $1$ ($M$ must be a closed interval).
** Indeed, since $f(\partial M) \subseteq \partial M$ and it's easy to see that $f(M^o) \subseteq M^o$ we get that $f(M^o)$ is clopen in $M^o$, thus $f(M^o)=M^o$. Since $f(M)$ is closed in $M$, and contains $M^o$, we conclude $f(M)=M$, and so $f(\partial M)= \partial M,f(M^o)=M^o$.
So, $f$ is a surjective $1$-Lipschitz map from a compact space to itself. (The $1$-Lipshictzity follows since isometric immersions preserve lengths of paths). Thus, it is a metric isometry. Hence, by the positive answer to this question, $f$ is smooth, and in fact a Riemannian isometry.
Best Answer
Let me mimic the argument of Całka:
Note that the map $f$ is short. Set $$X=\bigcap_n\, f^n(M)$$ Note that $f|_X$ is an isometry.
Note that for any positive integer $n$, any small convex ball in the interior of $M$ is mapped by $f^n$ isometrically. In particular $X$ has nonempty interior.
Indeed, a $2{\cdot}r$-ball in the interior is mapped to a $2{\cdot} r$-ball and the map is length-preserving. It follows that the corresponding $r$-ball maps isometrically to an $r$-ball.
Assume $X\ne M$. Fix $z\in M\backslash X$ which lies in a small ball centered in $X$. Note that the distance from $f^n(z)$ to $X$ is constant. On the other hand the set of partial limits of $f^n(z)$ have to lie in $X$, a contradiction.
So, $f$ is an isometry as short map from a compact space to it self. In particular $f(\partial M)=\partial M$