Does a unipotent transformation preserve covolume? Let $S\subset \mathbb{R}^d$ and define $v(S)$ to be the volume of the set $(S+\mathbb{Z}^d)\cap [0,1]^d$ (where $[0,1]^d$ is the unit cube $[0,1]\times [0,1] \times...\times [0,1]$).  Let $T: \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a linear transformation of determinant one whose only eigenvalue is one.  Is it necessarily true that $v(S)=v(T(S))$?

Let $S=[0,1/2]\times[0,2]$ and $T=\begin{bmatrix}1&1/2\\0&1\end{bmatrix}$. Then $v(S)=1/2$ but $v(T(S))=1$.
