This problem encountered me during the study of balanced Boolean functions. I was in need of constructing those balanced Boolean functions for which $\mathcal{D}_bf\ne 1$. And finding the solution to this problem got converted to finding the solution of this problem.
Let $n\geq 4$ and $\mathcal{A}\subseteq \mathbb{F}^{n}_{2}$ with $\lvert\mathcal{A}\rvert=2^{n-1}$. Let matrix$(\mathcal{A})$ be the $2^{n-1}\times n$ sized matrix obtained by assuming elements of $\mathcal{A}$ as rows. Let $set(\mathcal{B})$ be set of all the rows of matrix $\mathcal{B}$. For $\textbf{b}\in \mathbb{F}^{n}_{2}, \textbf{b}+\mathcal{A}:=\{\textbf{b}+\textbf{a} : \textbf{a}\in \mathcal{A}\}.$
$$\underline{\text{The problem is stated as follows:}}$$
Suppose $n\geq 4$ and $\mathcal{A}\subseteq \mathbb{F}^{n}_{2}$ with $\lvert\mathcal{A}\rvert=2^{n-1}$ and for every $\textbf{b}\in \mathbb{F}^{n}_{2},$ we have $\mathcal{A}\cap (\textbf{b}+\mathcal{A})\neq \emptyset.$
$\mathit{Q1.}$ Find the number of such sets $\mathcal{A}$.
$\mathit{Q2.}$ For a given such set $\mathcal{A}$, find the number of $\mathcal{H}\in \mathcal{GL}(n, \mathbb{F}_{2})$ (where $\mathcal{GL}(n, \mathbb{F}_{2})$ is general linear group of $n*n$ matrices over $\mathbb{F}_{2}$) such that $set(matrix(\mathcal{A})*\mathcal{H})=\mathcal{A}.$
$$\underline{Example:}$$ For $n=4.$ There are $10080$ such sets $\mathcal{A}$. $\text{If}\ \mathcal {A} = \{[0 0 0 0],[1 0 0 0],[0 1 0 0],[1 1 0 0],[0 0 1 0],[1 0 1 0],[0 0 0 1],[0 1 0 1]\},$ equivalently, $$matrix(\mathcal{A})=\begin{pmatrix} 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 1& 0& 0\\ 0& 0& 1& 0\\ 1& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 1 \end{pmatrix},$$ We get $8$ matrices from $\mathcal{GL}(4, \mathbb{F}_{2}):$
$\mathcal{H}_{1}=$ $\begin{pmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ \end{pmatrix}$, $\mathcal{H}_{2}=$ $\begin{pmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 0& 1& 0\\ 0& 1& 0& 1\\ \end{pmatrix}$, $\mathcal{H}_{3}=$ $\begin{pmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ \end{pmatrix},$
$\mathcal{H}_{4}=$ $\begin{pmatrix} 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ \end{pmatrix}$,
$\mathcal{H}_{5}=$ $\begin{pmatrix} 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 1& 0& 1& 0\\ \end{pmatrix}$, $\mathcal{H}_{6}=$ $\begin{pmatrix} 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ \end{pmatrix}$, $\mathcal{H}_{7}=$ $\begin{pmatrix} 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 1\\ 1& 0& 1& 0\\ \end{pmatrix},$
$\mathcal{H}_{8}=$ $\begin{pmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 0& 1& 0\\ 0& 0& 0& 1\\ \end{pmatrix}$
These are the precise matrices for which $set(matrix(\mathcal{A})*\mathcal{H}_i )=\mathcal{A}.$ Please help me in this regard. Thanks in advance.