I would like to characterize the set of all continuous $G:[0,1]^n \rightarrow [0,1]^n$ such that the Jacobian determinant is one:
$\mathcal{G}=\{G:[0,1]^n \rightarrow [0,1]^n: |DG(x)|=1\forall x \}$.
Is it true that $\mathcal{G}$ corresponds to the set of all linear transformations $G(x)=Ax+b$ with orthogonal matrices $A$? Indeed we have $DG= A$ in this case. And taking a determinant equal to one should imply that $A$ is orthogonal.
First, you need your functions to be $C^1$ (or at least have 1st order partial derivatives exist) for the Jacobian to be defined.
Second, $(-1)$ is an orthogonal $1 \times 1$ matrix, but $G(x) = -x$ does not define a function $[0,1] \to [0,1]$, so you conjecture is false for that reason.
Why do you believe that there is a nice classification of such maps?
Third, I would note that you really only want to be talking about differentiability on open sets, though as long as you want continuous differentiability, I think you'll be fine.
Fourth, I can't think of any nontrivial (as in not the identity map) functions with Jacobian 1 that map $I^n \to I^n$. Can you give an example? There might not be any.