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.

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.

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.