$2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns I am interested in answering the following question:
Question
For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column?
Conjectures

*

*For $n=2^k$, $k\ge0$ such a matrix always exists.


*For $n=3$ such a matrix does not exist.


*For $n=5$ such a matrix exists, for example:
$$
\left(
\begin{array}{cccccccccc}
 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\
 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\
 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\
 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\
 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\
 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\
 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\
\end{array}
\right)
$$


*For $n=7$ such a matrix does not exist.

Your first conjecture was proven by Nate in the comments.
Your second conjecture is also true - there is no such matrix for $n=3$. If we just look at which entries are nonzero in each row, because any two rows are orthogonal, they must share an even number of nonzero entries, i.e. either share $0$ entries or $2$ entries. Only $3$ other rows can share $2$ nonzero entries with a given row, since otherwise there'd be more than $3$ nonzero entries in a column by the pidgeonhole principle, so $2$ rows must share $0$ nonzero entries, but then they share $3$ entries with each other, contradiction.
