I have a $m\times m$ symmetric matrix $\{x_{ij}\}_{i,j\in I}$. We now partition the index set $I = \{1,...,m\}$. Let $S$ be the set of partitions of the index set . We denote any partition in $S$ by $[k]$, where $k$ is some element in $[k]$. Clearly for any $S$, $\bigcup_{{[k]}\in S} [k] = I$. Now, I have the following conditions, for all $[k]$ :
$$\forall p,q \in I: \{x_{pi}: i\in [k] \} = \{x_{qi'}: i'\in [k] \}$$
Now, it is easy to infer that if all the $[k]$ are singletons then all the $x_{ij}$ are the same number. On the other hand if $[k] = I$ then all the rows are the same set of elements, but are a permutation of one another. I am confused about what happens in between ? i.e. there are many $[k]$ larger than a singleton. Can I derive a simple condition which encompasses all $x_{ij}$? Are such matrices known ?
I think the resultant condition is that the matrix $\{x_{ij}\}$ is a block matrix, with each block being the same symmetric matrix. I am not sure how to prove it yet though !
I found a symmetical matrix that works for $m = 5$, $S = \{\{1,2,3\},\{4,5\}\}$.
$$X = \begin{bmatrix}0&1&1&0&1 \\ 1&0&0&1&0 \\ 1&0&1&0&1 \\ 0&1&0&1&0 \\ 1&0&1&0&1\end{bmatrix}$$
Here, each row in the first $3$ columns always contain all the numbers in $\{0,1\}$, and each row in the last $2$ columns also always contain $\{0,1\}$.
I feel like this has to be possible for any $m$ and $S$, but I couldn't find a way to fill the matrix with the values $\{0,1,2\}$. So you just have to be careful to keep the symmetry, and you might not be able to use $\max_{[k] \in S} |[k]|$ numbers (which is $3$ in this case).