Cross posting from MSE, where [this question](https://math.stackexchange.com/q/4369960/643521) received no answers.


The following Latin square


$$\begin{bmatrix}
1&2&3&4&5&6&7&8\\
2&1&4&5&6&7&8&3\\
3&4&1&6&2&8&5&7\\
4&3&2&8&7&1&6&5\\
5&6&7&1&8&4&3&2\\
6&5&8&7&3&2&4&1\\
7&8&5&2&4&3&1&6\\
8&7&6&3&1&5&2&4
\end{bmatrix}$$


has the property that for all pairs of two different rows $a$ and $b$, the permutations $ab^{-1}$ have the same cycle type (one 2-cycle and one 6-cycle).


What is known about Latin squares with the property that all $ab^{-1}$ have the same cycle type (where $a$ and $b$ are different rows)? For example, do they have a particular structure, for which cycle types do they exist, are there any infinite families known, do they have a name, etc?


The only example of an infinite family I'm aware of are powers of a single cyclic permutation when $n$ is prime, for example:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&4&5&1\\
3&4&5&1&2\\
4&5&1&2&3\\
5&1&2&3&4
\end{bmatrix}$$



Latin squares with one cycle type?
One way to achieve the required property is to construct a Latin square whose autotopism group acts transitively on unordered pairs of rows. This can be achieved for orders that are a prime power congruent to 3 mod 4, by means of the quadratic orthomorphism method, described [in this paper](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v12i1r22). The focus of that paper is the atomic squares I mentioned in other comments, but the construction will always achieve the property that each pair of rows produces the same cycle structure. For prime powers that are 1 mod 4, quadratic orthomorphisms will build Latin squares with at most two types of cycle structures formed by pairs of rows.


