Determine a matrix from terms of Leibniz formula Let $A$ is an $n\times n$ matrix with nonzero real entries $a_{ij}$.  Say one is given
$$
X_{\sigma}:=\text{sgn}(\sigma)\cdot\prod_{i=1}^na_{i\,\sigma(i)}
$$
for each $\sigma\in S_n$.  Is this information enough to determine $A$?  I believe the answer is no by the following argument:
One can `linearize' this system by taking logs and setting $b_{ij}=\log a_{ij}$.  Then one notes that the space of permutation matrices has dimension $(n-1)^2+1$, but there are $n^2$ variables $b_{ij}$, so the space of solutions has dimension $n^2-(n-1)^2-1$.  However, this would seem to imply that the linear map $[b_{ij}]_{1\le i,j\le n}\mapsto [X_\sigma]_{\sigma\in S_n}$ is not injective, but I don't see plainly what the kernel is?
I'm also interested in the modification where we only see $X_\sigma$ if $\sigma$ has at least one fixed point, and we just want to find $\det A$.
Edit: I'm interested in the case where the number of constraints $\lceil n!(1-1/e)\rfloor$ exceeds the number of variables $n^2$
 A: For $n > 1$, if $E$ is the set of $n{\,\times\,}n$ real matrices with at most $n-1$ nonzero entries, then for all $A\in E$, we have $X_{\sigma}=0$ for all $\sigma\in S_n$, so $A$ is not uniquely determined.

For another example, for $n > 1$, let $A$ be an $n{\,\times\,}n$ real matrix with all rows equal but no two columns equal, and let $B$ be any matrix obtained from $A$ by a non-identity permutation of the columns of $A$. Then $A\ne B$, but for all $\sigma\in S_n$ the value of $X_\sigma$ for $A$ is equal to the value of $X_\sigma$ for $B$, so the given information is not enough to separate $A$ and $B$.

For the modified question, let $D$ be the set of $2{\,\times\,}2$ real matrices with at least one zero on the main diagonal. Noting that the identity permutation is the only element of $S_2$ with at least one fixed point, it follows that the known information is the same for all $A\in D$, but $\det(A)$ is not uniquely determined.

Update:

My above answer applies to the OP's question as originally stated.

But for the OP's current version of question $(1)$, we can still show non-uniqueness . . .

Fix $n > 1$.

Let $A$ be an $n{\,\times\,}n$ real matrix all of whose entries are nonzero.

Let $x_1,...,x_n,y_1,...,y_n$ be $2n$ real numbers whose product is $1$ and such that $x_iy_j\ne 1$ for at least one pair $(i,j)$.

Let $B$ be the $n{\,\times\,}n$ real matrix such that for all $i,j$ we have $b_{ij}=x_iy_ja_{ij}$.

Then $A\ne B$, but for all $\sigma\in S_n$ the value of $X_\sigma$ for $A$ is equal to the value of $X_\sigma$ for $B$, so the given information is not enough to separate $A$ and $B$.
