Determinantally equivalent nonzero functions

We study the problem raised in [Marco Stevens, Equivalent symmetric kernels of determinantal point processes, RMTA, 10(03):2150027, 2021] concerning the extension of its main result to the more general (potentially non-symmetric) setting. We construct a counterexample disproving the conjecture proposed in the paper, and subsequently solve it under some additional minor assumptions that preclude such counterexamples. The problem is plainly stated as follows: Let $\Lambda$ be a set and $\mathbb{F}$ a field, and suppose that $K,Q:\Lambda^2\to\mathbb{F}$ are two functions such that for any $n\in\mathbb{N}$ and $x_1,x_2,\ldots,x_n\in\Lambda$, the determinants of matrices $(K(x_i,x_j))_{1\leq i,j\leq n}$ and $(Q(x_i,x_j))_{1\leq i,j\leq n}$ agree. What are all the possible transformations that transform $Q$ into $K$? In [Marco Stevens, Equivalent symmetric kernels of determinantal point processes, RMTA, 10(03):2150027, 2021] the following two were conjectured: $(Tf)(x,y)=f(y,x)$; and $(Tf)(x,y)=g(x)g(y)^{-1}f(x,y)$ for some nowhere-zero function $g$. In the same paper, this conjectured classification is verified in the case of symmetric functions $K$ and $Q$. By extending the graph-theoretic techniques of the paper, we show that under some surprisingly simple and natural conditions the conjecture remains valid even with the symmetry constraints relaxed. By taking $\Lambda$ finite, the above problem, furthermore, reduces to that between two square matrices investigated in [Raphael Loewy, Principal minors and diagonal similarity of matrices, Linear Algebra and its Applications 78 (1986), 23--64]. Hence, our paper presents a simple non-linear-algebraic proof that uses only some elementary combinatorics and three simple algebraic identities involving $3$-cycles and $4$-cycles.