Exotic homeomorphisms of a cube

Question

If $\varphi:\mathbb{R}\to\mathbb{R}$ is continuous, non-constant, non-decreasing, and differentiable a.e. with $\varphi'=0$ a.e., then the mapping $$ \Phi(x,y)=(x+\varphi(x),y+\varphi(y)) $$ is a homeomorphism $\Phi:\mathbb{R}^2\to\mathbb{R}^2$ that is differentiable a.e., $D\Phi=I$ a.e., but $\Phi$ is not a translation. That shows there are non-trivial homeomorphisms of $\mathbb{R}^2$ whose a.e. derivative is the identity. Clearly, similar examples exist in any dimension.

Question. Let $n\geq 2$ and $Q=[0,1]^n$. Suppose that $\Phi:Q\to Q$ is a homeomorphism such that $\Phi|_{\partial Q}=\operatorname{id}$, $\Phi$ is differentiable a.e. and $D\Phi=I$ a.e. Does it follows that $\Phi(x)=x$ for al $x\in Q$?

When $n=1$ i.e. the answer is yes, because it is easy to show that $\Phi$ is absolutely continuous.

Answer 1

The answer is no and a suitable counterexample in any dimension $n\geq 2$ was constructed by Oxtoby and Ulam. This is Theorem 12 p. 919 in:

J. C. Oxtoby, S. M. Ulam, Measure-preserving homeomorphisms and metrical transitivity. Ann. of Math. 42 (1941), 874–920.

Here, the automorphism means homeomorphism. Since it is rigid translation on cubes, the derivative is $DT^*=I$ on each of the cubes. This theorem says that not only strange homeomorphisms answering the question exist, but every homeomorphism preserving the measure can be uniformly approximated by such homeomorphisms. This is really a quite amazing result.