Given $M$ a symmetric matrix in $\mathbb F\_2^{n\times n}$ having $\mathsf{det}\_\mathbb R(M)\neq0$ (*non-singular in reals*) and satisfying $PMP'=(M+J+I)$ or $P(M+J+I)P'=M$ where $P$ is a permutation matrix in $\mathbb F\_2^{n\times n}$ and where $J$ is all $1$s and $I$ is identity matrix in $\mathbb F\_2^{n\times n}$ ($'$ is transpose and so $P'=P^{-1}$ holds).


$\mathsf{Det}(M)=\mathsf{Det}(M+J+I)$ and $\mathsf{Det}(M+I)=\mathsf{Det}(M+J)$ are satisfied as $\mathsf{Det}(P)\equiv1\bmod2$.



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> Can $M,M+I\not\in\mathsf{SL}(n,\mathbb F\_2)$ be simultaneously impossible?
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> Is there an example of $\mathsf{det}(M)+\mathsf{det}(M+I)\equiv1\bmod2$?
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*Update* Antoine Labelle's comments below suggests either $M,M+I\in\mathsf{SL}(n,\mathbb F\_2)$ or $M,M+I\not\in\mathsf{SL}(n,\mathbb F\_2)$ holds and he has no situation $\mathsf{det}(M)+\mathsf{det}(M+I)\equiv1\bmod2$.



On a matrix problem in the field $\mathbb F_2$
Most of the question has already been addressed. As regards the last part, I just poit out that if ${\rm det} M = 0$ and ${\rm det}(M+I) = 1$, then we must have ${\rm rank}(M) = n-1$ ( I mean here the $\mathbb{F}\_{2}$ rank). This follows since $M + I + J$ is (by assumption) similar to $M$, and hence has the same rank. If that rank is $n-2$ or less, then the null space of $M+I+J$ is at least two-dimensional, and the null space of $J$ is $n-1$ dimensional. Hence there is a non-zero column vector $v$ with $(M+I+J)v = 0$ and $Jv = 0.$ Hence $(M+I)v = 0,$ so that ${\rm det}(M+I) = 0.$


