Upper bound on number of pairwise 'coupled' 'full' matrices of order $n$ Let us define a 'full' matrix of order $n$ as a square array of numbers such that each row and column contains all the numbers from 1 to $n$. Given such a matrix $A$ we indicate the element in the $i^{th}$ row and $j^{th}$ column as $a_{ij}$. Furthermore, we say two such matrices $A, B$ of order $n \geq 3$ are 'coupled' if all the pairs $(a_{ij}, b_{ij})$ are distinct for all $1\leq i, j \leq n$.
For example, taking $n=3$, the following two matrices are 'coupled'
\begin{bmatrix}
1 & 2 & 3\\
2 & 3 & 1 \\
3 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 3\\
3 & 1 & 2 \\
2 & 3 & 1
\end{bmatrix}
as the generated pairs are respectively $(1, 1), (2, 2), (3, 3), (2, 3), (3, 1), (1, 2), (3, 2), (1, 3), (2, 1)$.
The question is to find an upper bound $c(n)$ for the number of 'full' matrices of order $n$ that are pairwise 'coupled'.
By testing small $n$ by hand I have come to the conjecture that $c(n) = n-1$, but I am not able to prove it. Does anyone have any ideas?
 A: You are correct that $n-1$ is an upper bound. Here is an outline of a proof, with gaps for you to fill in.

*

*Given a full matrix $A$, and a permutation $\pi$ of $\{1,\dots,n\}$, you can form a new full matrix $\pi(A)$ by applying $\pi$ to each entry of $A$. First, you should prove that if $A$ and $B$ are coupled, then $\pi(A)$ and $B$ are coupled as well.


*Secondly, use the previous step to show that every family of pairwise coupled full matrices can be put in "standard form," meaning the first row of each matrix is $[1,2,\dots,n]$ in that order.


*Once the family $A_1,A_2,\dots,A_k$ is in standard form, let $x_i$ denote the entry in the second row, first column of $A_i$. Every matrix in the family looks like this: $$
A_i=\begin{bmatrix}
1&2&3&\cdots&n\\
x_i &&&\dots\\
\vdots&&&\ddots
\end{bmatrix}
$$
What must be true about the numbers $x_1,x_2,\dots,x_k$? Use this to conclude $k\le n-1$.
Furthermore, you can show that this upper bound cab ne obtained whenever $n$ is a prime power. If $n$ is prime, this is not too hard to do using matrices defined with arithmetic modulo that prime. When $n$ is a prime power, you instead have to use finite field arithmetic.

In case you are not aware, a "full matrix" is usually called a Latin square, "coupled" matrices are usually called orthongal, and a family of pairwise coupled full matrix is called an MOLS, short for mutually orthogonal Latin squares.
