Partial order over a set of matrices Let $\textbf{S}$ be the set of square matrices of order $n$ where the entries $a_{ij}$ are integers such that $1 \le a_{ij} \le n$
Let $\textbf{G}$ be the group of transformations acting on $\textbf{S}$, generated by swap of rows, swap of columns and symmetry with respect to the main diagonal.
$\textbf{G}$ is of order $2(n!)^2$
Two matrices $Q$ and $R$ of $ \textbf{S}$ are equivalent if there exists $g \in \textbf{G}$ such that $Q=g(R)$
Note:
$Q =\{q_{ij}\}$
$R =\{r_{ij}\}$
Define the binary relation $\cal \preceq$ on $\textbf{S}$:
$Q\cal \preceq$ $ R$ if it exists $R'=\{r'_{ij} \}$ equivalent to $R$, such that:
$\qquad q_{ij} \le r'_{ij} \;\;  \forall i : 1\le\ i \le n $ and $\;\forall j : 1\le\ j \le n $
Is the binary relation $\cal \preceq$ a partial order on $\textbf{S}$?
If it is true, is the property valid for any group $\textbf{G}$ acting on $\textbf{S}$?
 A: The relation is a partial order on $S$ for any group $G$ that permutes the elements of a matrix, but does not change their values.
(If $G$ is allowed to make arbitrary changes to the elements of the matrix, then even in the case of $n=1$, we can find a $G$ that does not give a partial order: let $G = \mathbb Z^+$, and for a matrix $(x) \in S$ let $g\in G$ act on $(x)$ to give $(x + 2g)$.)
Proving reflexivity is easy. Antisymmetry is nearly as easy, though it is the property that fails for the example given above. The tricky one is transitivity.
Let $q, r, s\in S$ be $n\times n$ matrices. Suppose $q\leq r$ and $r \leq s$. In other words, there are elements $g_1, g_2 \in G$ such that $\forall i,j; q_{ij}\leq (g_1r)_{ij}$ and $\forall i,j; r_{ij}\leq (g_2 s)_{ij}$. Because the group action can only permute matrix elements, not change their values, we can apply $g_1$ to both sides of the second expression to obtain: $(g_1r)_{ij} \leq (g_1g_2s)_{ij}$. Then we have $\forall i,j; q_{ij}\leq (g_1r)_ij \leq (g_1g_2s)_{ij}$. Therefore $\forall i,j; q_{ij}\leq (g_1g_2 s)_{ij}$, which implies that $q\leq s$, as required.
