Suppose $f,g: U \subset \mathbb{R}^n \to \mathbb{R}^n$ are smooth immersions from a closed and bounded domain $U$ such that $f|_{\partial U} = g|_{\partial U}$. If $f$ and $g$ induce the same Riemannian metric on $U$ (i.e. $\langle df, df \rangle = \langle dg, dg \rangle$), does it follow that $f = g$ pointwise?
I suspect the answer is yes, and I have a sketch of a proof for $n=2$. In particular, the hypotheses imply that (at least locally), we have $(g \circ f^{-1})^*\delta = \delta$ where $\delta$ is the Euclidean metric on $\mathbb{R}^2$. Then, $g\circ f^{-1}$ is identity on the boundary, and to show that it is also identity on the interior we can consider any Euclidean triangle in $f(U)$ with two vertices fixed on (some convex part of) the boundary and the third in the interior. This triangle has two vertices and all edge lengths preserved under $g\circ f^{-1}$, therefore its third vertex is preserved also. If it is not clear enough from this, I think starting close to the boundary and ''bootstrapping'' the same procedure toward the interior of the domain should allow it to be used anywhere, but I have not thought carefully about it yet.
I know that codimension-zero immersions are quite rigid as a general rule, but so far I cannot come up with a proof nor a counterexample for this statement. Do you know of any arguments or references which might answer this question?
With the clarification of the definition, here is a proof:
Theorem. Suppose that $M$ is a smooth $n$-dimensional compact connected manifold with boundary, $f, h: M\to {\mathbb R}^n$ are immersions, i.e. differentials $df_x, dh_x$ have rank $n$ everywhere in $M$, such that $f|\partial M= h|\partial M$ and $g=f^*(g_0)=h^*(g_0)$ where $g_0$ is the standard flat metric on ${\mathbb R}^n$. Then $f=h$.
Proof. Consider a point $x$ in the interior of $M$. Pick a unit tangent vector $u\in T_xM$ and let $c$ denote the maximal unit speed geodesic segment in $(M,g)$ such that $c(0)=x, c'(0)=u$. Maximality here mean that the domain $J$ of $c$ is maximal possible subinterval in ${\mathbb R}$. Geodesic is understood in the Riemannian sense, i.e. $\nabla_{c'}c'=0$, where $\nabla$ is the LC-connection of $g$.
Lemma. $J$ is a compact interval $[a,b]$ such that $c(a), c(b)\in \partial M$.
Proof. There are several cases one has to exclude, I will eliminate the most interesting one, when $J$ is unbounded. Observe that $f\circ c: J\to {\mathbb R}^n$ has image contained in $f(M)$, which is a compact subset of ${\mathbb R}^n$. But $f\circ c$ is a parameterized with the unit speed infinite Euclidean line segment. This is a contradiction. I will leave it to you to complete the proof in the case when $J$ has the form $(a,b), (a,b], [a,b)$. (You again use compactness of $M$ and the fact that the metric $g$ is defined in the entire $M$, including its boundary.) qed
Thus, we have, $f\circ c, h\circ c: [a,b]\to {\mathbb R}^n$ are isometric maps, which, by the hypothesis of the proposition, satisfy $fc(a)=hc(a), fc(b)=hc(b)$. Now, it is a simple fact of Euclidean geometry that $f\circ c= h\circ c$. In particular, $f(x)=h(x)$. qed