order of a permutation and lexicographic order Let $M$ be an $n\times m$ matrix, say with entries in  $\left\{0,1\right\}$ ; and  let $\mathcal C(M)$ be the $n\times m$ matrix such that there exists $P$, $m\times m$ permutation  matrix such that $M.P=\mathcal C(M)$ and such that the columns of $\mathcal C(M)$ are lexicographically increasing (1) (for a formal definition of (1) see reflexive relations that are "tridiagonally cycle-indexed" (or "almost ordered" matrices/relations))
$\mathcal R(M):= \mathcal C(M^t))^t$ is the matrix you get from $M$ that rows are lexicographically increasing. 
We now say that $\mathcal L=\mathcal C\,o\,\mathcal R$.
Let $Q$ be a $m\times m$ permutation matrix s.t. $Q^q=Id$. We  define $\mathcal L_Q$ to be such that $\mathcal L_Q(M)=\mathcal L(M).Q$ for all $M$ of size $n\times m$. 

Does there exists $r\in \mathbb N$ such that $\mathcal L_Q^{r+iq}(M)=\mathcal L_Q^r(M)$ for any  $i\in \mathbb N$

The cases that seem the most interesting to me are $\mathcal L_{Id}=\mathcal L$ and $\mathcal L_J$ where $J$ is the $i\mapsto m-i$ permutation matrix, I talked about these cases in the upper link.

Example with $m=n=4$ 
$Q=J=\begin{matrix}
0 & 0 & 0 & 1  \\ 0 & 0 & 1 & 0  \\ 0 & 1 & 0 & 0  \\ 1& 0 & 0 & 0  
\end{matrix}$
(so $q=2$)
Let's take
 $M=\begin{matrix}
0 & 0 & 1 & 1  \\ 1 & 0 & 0 & 1  \\ 0 & 1 & 1 & 0  \\ 1& 1 & 1 & 0  
\end{matrix}$
We range rows according to lexicographic order :
$\mathcal R(M)=\begin{matrix}
0 & 0 & 1 & 1\\ 0 & 1 & 1 & 0  \\ 1 & 0 & 0 & 1    \\ 1 & 1 & 1 & 0  
\end{matrix}$
And now columns...
$\mathcal L(M)=\begin{matrix} 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 1  \\ 1 & 0 & 1 & 0  \\ 1 & 1 & 0 & 1  
\end{matrix}$
And we multiply on the right by $J$:
$\mathcal L_J(M)= \begin{matrix} 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0  \\ 0 & 1 & 0 & 1  \\ 1 & 0 & 1 & 1  
\end{matrix}$
If we apply $\mathcal L_J$ to $\mathcal L_J(M)$ we then get :
$\mathcal L^2_J(M)= \begin{matrix} 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1  \\ 1 & 0 & 1 & 1  \\ 0 & 1 & 1 & 0  
\end{matrix}$
etc....
We will get :
$\mathcal L^4_J(M)=\mathcal L^6_J(M)=\begin{matrix} 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0  \\ 0 & 0 & 1 & 1  \\ 0 & 1 & 1 & 1  
\end{matrix}$
We could verify that $\mathcal  L^3_J(M)\ne \mathcal L^5_J(M)(\ne L^4_J(M))$ and then $r=4$ is the smallest possible (and $k$ such that $i\mapsto L^{r+i}_J(M)$ is $2$-periodic but not constant)

There are counterexamples for $Q=J$. 
Here is a binary $6\times7$ binary matrix $M$ that belongs to an orbit of $\mathcal{L}_J$ with period $3$:
$\begin{matrix}
1&1&1&0&0&0&0\\
1&1&0&1&1&0&0\\
0&0&1&1&0&1&0\\
0&0&1&0&1&0&1\\
1&0&0&1&0&1&1\\
0&1&0&0&1&1&1\\
\end{matrix}$
Here is another with period $4$:
$\begin{matrix}
1&1&0&0&0&0&0\\
1&0&1&1&0&0&0\\
0&1&1&0&1&0&0\\
0&1&0&1&0&1&1\\
1&0&0&0&1&1&0\\
0&0&0&1&1&1&0\\
\end{matrix}$
and another with period $6$:
$\begin{matrix}
1&1&1&0&0&0&0\\
1&1&0&1&1&0&0\\
0&0&1&1&0&1&0\\
0&0&1&1&0&0&1\\
1&0&0&0&0&1&1\\
0&1&0&0&0&1&1\\
\end{matrix}$
On the whole set of $7\times 7$ binary matrices, $\mathcal{L}_J$ gets
$$
\begin{array}{r l}
     326\,166&\text{fixed matrices}\\
 86\,146\,036&\text{distinct orbits of length }2\\
           94&\text{distinct orbits of length }3\\
       5\,400&\text{distinct orbits of length }4\\
            8&\text{distinct orbits of length }5\\
          196&\text{distinct orbits of length }6\\
\end{array}
$$
