Hi, originally I was wondering if there exists a set of (equivalent) Latin squares that are closed under matrix multiplication. The conditions on the permutations come from the latin square constraints. The last one from the latin squares matrix multiplication clousure condition. Cheers.

Thanks for your answer! I was looking indeed for some simple characterization but could not find it. Here was my (not conclusive) reasoning: if L is a latin square I can write it as: $\sum_i^k P_i v_i$ for some set of permutations $P_i$ such that $\sum _i P_i =J$ the all-ones matrix and $v\in R^k$. Thus using a similar reasoning for $L'$ I can write $\sum_i^k A^TP_iA v_i=\sum_i \tilde{P}_ir_i$ for some set of permutations $\tilde{P}_i$ and $r\in R^k$. From this i wanted to deduce that $A^TP_iA $ must be a permutation and thus $A$ must be a permutation and $v$ proportional $r$...

Hi, I guess in that case P=e the identity,which is a permutation

For example the function $f(\lambda)=(1/N)H(\cdot-\lambda)$ ($H$ the Heaviside) will work. This correspond to the cumulative distribution function of the set $S$. As a side comment I found out that the function $(1/N)\sum_{i}^{N}H(x_i-\lambda)$ is a maximal invariant w.r.t. the permutation group (of order $N$) and any bijective function of it is still a maximally invariant.

Calling the moments $c(p)=\sum_{i}^{N}x^p_i$, the question may be rephrased as: which (finite or not) combinations of moments $f(\lambda,x_i) = \sum_{p} \lambda(p)x_{i}^{p}$ is such that the set $q(\lambda)=\sum_{i}^{N}f(\lambda,x_{i})$ is still characterizing $S$? (If I am not wrong this is called the generalized moment problem )

Thanks for all the answers! The moments observation is interesting. In fact another way to characterize the sequence $S$ is to have all its moments

Mark, you are right! probably nonlinearity is playing a crucial role here and $f$ should be nonlinear monotonic.

Sorry I should heve been more clear. I corrected the question above. Thanks !

@Survit Hi Survit, I can rewrite it as $\sum_{p=1}^{\infty} ||A^{\circ p}J- JA^{\circ p} ||_{F}^{2}$. Thanks.

@GeoffRobinson Hi Geoff, the matrix A should be coming from a multiplication table of a group.