Consider $\mathbf{A}$, an $n \times n$ matrix over $\mathbb{F}_q$, the finite field with $q$ elements. The transpose of $\mathbf{A}$ is denoted with $\mathbf{A}^T$. Let $\mathbf{I}_n$ denote the identity matrix of order $n$. Assume that the matrix $\bf A$ has the following property over $\mathbb{F}_q$, $ \mathbf{A}\mathbf{A}^T=\mathbf{I}_n. $
We say $\mathbf{B}$ is a linear combination of the set $\{\mathbf{A},\mathbf{A}^T\}$, if $\mathbf{B}=\alpha_1\mathbf{A}+\alpha_2\mathbf{A}^T$ where $\alpha_i$ with $1\leq i \leq 2$ are elements in $\mathbb{F}_q$.
My Question: How to find an $n\times n$ matrix $\mathbf{B}$ such that $\mathbf{B}\mathbf{B}^T=-\mathbf{I}_n$ over $\mathbb{F}_q$ whithout restriction on $q$?
Example: Consider $\alpha \in \mathbb{F}_q$ such that $\alpha^2=-1$, then by considering $\mathbf{B}=\alpha\mathbf{A}$, we get $\mathbf{B}\mathbf{B}^T=-\mathbf{I}_n$. But due to the condition over $\alpha$, we have $q \equiv1\pmod{4}$.
This isn't always possible. For example, if $A=I$, then the problem collapses to $(\alpha A)(\alpha A)^T=\alpha^2 I=-I$, but $\alpha^2=-1$ does not always have solutions.
However, if $B=\alpha_1 A + \alpha_2 A^T$, then $BB^T=(\alpha_1^2+\alpha_2^2)I+\alpha_1 \alpha_2 (A^2+(A^T)^2)$. If $\alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.