Let $n\in\mathbb{N}_{\geq 1}$ be given.
Consider the vector space $\mathbb{C}^n$ and its standard basis $\{e_j\}_{j=1,\dots,n}$. Let $U$ be a unitary matrix, with columns $u_1,\dots,u_n$, and regard it as a map $$ U:\mathbb{C}^n\to\mathbb{C}^n\,. $$ I am interested in describing the space (as a set, or a $\mathbb{Z}$-module) $$ S := U(\mathbb{Z}^n) \equiv \{v \in \mathbb{C}^n | v = Ur\exists r\in\mathbb{Z}^n\}\,.$$
In particular as such it is clear that $S$ is not a cylinder set. However, it seems like it could be a union of cylinder sets.
In the simplest example, consider the case $n=2$ and the unitary matrix $$ U = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\,. $$
In this case it is easy to see that $$ S = (\sqrt{2}\mathbb{Z})^2\bigcup((\sqrt{2}\mathbb{Z})^2+\begin{bmatrix}1 \\ 1\end{bmatrix}) \,. $$
The reason is essentially that the pair of sum and difference of two integers always have the same parity, so while we can't just take the whole of $\mathbb{Z}^2$ to span these points, we can take $$ (2\mathbb{Z})^2 \bigcup (2\mathbb{Z}+1)^2$$ which up to scaling of the lattice constant is the same as what was written above.
Can a similar statement be proven for the general case?
It seems to me like the answer should be something like the following: Let $B$ be the open subset of $\mathbb{C}^n$ given by the interior of the hyperbuce spanned by the vectors $\sqrt{n} e_j$ for $j=1,\dots,n$. Then $$ S\equiv U(\mathbb{Z}^n) = (\sqrt{n}\mathbb{Z})^n\cup\bigcup_{j:u_j\in B}((\sqrt{n}\mathbb{Z})^n+u_j)\,.$$
(I realize this cannot be true for all unitary matrices $U$ but I'm not sure what class I'd like to restrict to; perhaps only circulant matrices)
Is this correct? If yes, how to prove it? Also, how many vectors are in $B$? How to describe them?