This question is a follow on from the related question here.
A nonogram (or griddler), for this purpose, is an $m\times n$ blank rectangular grid with clues for each row and column.
A solution involves filling each square in the grid with a symbol $0$ or $1$ matching the clues.
A clue consists of a sequence of integers - for a row of length $m$, the integers might be $a,b,c,d$ with $a+b+c+d+3\le m$ (the number $3$ is one less than the number of clue integers). The clue means that there may be some zeros to the left, but that the leftmost block of ones contains $a$ ones, then there is at least one zero before the next block of exactly $b$ ones (etc). There is a clue for each row and column. The integers in the clues refer to all the ones in the relevant row or column. See the linked question for more detail.
In an answer to the linked problem, I observed that where $m$ and $n$ are even, there are two checkerboard patterns possible, and each generates the same clues. It is possible to put an (even x even) checkerboard in the corner of an array and fill the rest with zeros to get an indeterminate set of clues. If either dimension is $1$ the whole thing is determined by the clues.
The checkerboard pattern has the strong property that none of the values can be determined from the original clues. Is there any (odd x odd) or (odd x even) grid which has this same strong property. (filling a corner with a checkerboard leaves some completely determined rows and/or columns)?
$$\pmatrix{ 1&0&0\\0&1&0\\0&0&1}$$ and $$\pmatrix{ 0&0&1\\0&1&0\\1&0&0}$$ have the same row and column descriptions. I'm not sure if this meets your criteria or not, because I don't perfectly understand what is meant by "none of the values can be determined." Any $3\times3$ permutation matrix has one $1$ in each row and column, and so has the same description. It seems to me that given this description, you can't deduce the value in any cell, but I'm not sure that's what you mean.