Kräuter's rank bound conjecture for permanents of sign matrices

Let $\mathbb{F}$ be a field of zero characteristics. For positive integers $k,n$, let $M_{k,n}(\mathbb{F})$ be the set of $k$-by-$n$ matrices over $\mathbb{F}$, and let $M_{k,n}(\pm 1)$ be the subset whose entries are $\pm 1$. Write $M_n=M_{n,n}$, let $\operatorname{rk} A$ denote the rank of $A$, and let $\operatorname{per} A$ denote its permanent. For $0\leq l\leq k\leq n$, define $D_{(n,k,l)}=(d_{ij})\in M_{k,n}(\pm1)$ by

$$d_{ij}=\begin{cases} -1, & i=j\text{ and }j\in\{1,\ldots,l\},\\ 1, & \text{otherwise.} \end{cases}$$

When $n=k$, write $D_{(n,n,l)}=D_{(n,l)}$. Kräuter's conjecture. Let $A\in M_n(\pm1)$, where $n\geq5$ and $\operatorname{rk} A=r+1$ for some $r$ with $0\leq r\leq n-1$. Then

$$|\operatorname{per} A|\leq \operatorname{per}D_{(n,r)}.$$

Equality holds if and only if $A$ can be obtained from $D_{(n,r)}$ by transposition, row or column permutations, and multiplication of rows or columns by $-1$. This conjecture would answer Wang's question by giving a sharp upper bound for the permanent of a sign matrix in terms of its rank; the stated source does not establish the claim, and its resolution status is not supplied here.