Cardinality of the maximum points of the determinant on matrices with entries in [-1, 1] By multilinearity, the maximum of the determinant on matrices with entries in the interval [-1, 1] is attained at a {-1, 1}-matrix. By the following example, the maximum is attained at infinitely many points on matrices of order $3$
\begin{bmatrix}
1 & 1 & 1\\
-1 & 1 & 1\\
t & 1 & -1
\end{bmatrix}
$-1 \leq t \leq 1$.
However, I verified that there are only finitely many maximum points for orders $4, 5, 6$. So, I ask, is the maximum attained only at finitely many points for order above $3$, or perhaps for orders not equal to
$3$ mod 4? I suspect this is known, as there has been a lot of work on Hadamard Maximum determinant problem, but I can't find such statement so far.
If this is not known, can we at least prove that the dimension of the space of maximum points is less than the order?
 A: This response was too long for a comment, but is far from a complete answer.
Much of the research on the maximal determinant problem has focused on the Gram matrix, and used the theory of positive definite matrices to obtain bounds. For orders $n \equiv 0 \bmod 4$ the optimal Gram matrix is $nI_{n}$ and the maximal determinant matrices are Hadamard matrices. Since the entries must be $\pm 1$ and rows must be orthogonal, the set of solutions will be finite, and there will be no free parameters of the type in the question since these would appear in the Gram matrix, and you could apply Fischer's Inequality to push the determinant away from the Hadamard bound.
For orders $1, 2 \bmod 4$ there are Gram matrices which are known to be optimal, but these cannot always be attained for number theoretic reasons. When the optimal Gram matrices are known and attained, the number of solutions will be finite. For orders $3 \bmod 4$ there are not even conjectures for what the optimal Gram matrices should be. Forgive the self-promotion, but with some collaborators, I wrote a survey on the maximal determinant problem which might be of interest:
https://arxiv.org/abs/2104.06756
The free parameter you observe above occurs because the matrix has an $(n-1) \times (n-1)$ minor equal to $0$. This cannot occur in the Hadamard case, because the inverse matrix is proportional to the transpose. Outside of the Hadamard case, the inverse matrix doesn't figure directly in the maximal determinant problem, so I don't know that this problem has been considered. Hadamard matrices do contain many $(n-2)\times (n-2)$ minors equal to $0$, complementary to any vanishing $2\times 2$ minor, so the question may well be quite difficult in general.
Physicists have explored the existence of free parameters in complex Hadamard matrices, though they normally allow multiplication of a subset of entries in the matrix by a complex number of norm 1. Many examples are here:
https://chaos.if.uj.edu.pl/~karol/hadamard/
