Given an invertible $\{-1,0,1\}$-matrix $A$ (its determinant is $\pm 1$), are there two permutation matrices $P$ and $Q$ such that all the leading principal minors (determinants of the top-left sub-matrices) of $A' = P A Q$ are $\pm 1$?
For example, $$ A = \left[ \begin{array}{cccc} 0 & 1 & 1 & -1 \\ 0 & 0 & -1 & -1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 \end{array} \right], \quad \det(A) = 1 $$ and $$ A' = \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array} \right] \cdot A \cdot \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array} \right] = \left[ \begin{array}{cccc} -1 & -1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & -1 & 1 & 0 \end{array} \right] $$ accomplishes the condition.