Latin squares with one cycle type? Cross posting from MSE, where this question 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:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&4&5&1\\
3&4&5&1&2\\
4&5&1&2&3\\
5&1&2&3&4
\end{bmatrix}$$
 A: There are also "pan-Hamiltonian" Latin squares, see Perfect Factorisations of Bipartite Graphs and Latin Squares Without Proper Subrectangles by I. M. Wanless, Electronic J. Combin. 6 (1999), R9.
A: 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. 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.
A: Since the quasigroups are the algebraic structures that correspond to Latin squares, one of the first things that I would do is try to investigate if these Latin squares satisfy any special identities or if they can be characterized as the algebras that satisfy a collection of identities. It turns out that when we add the condition that all cycles in the permutations $x\mapsto a*(b\setminus x)$ are of length $p$ whenever $a\neq b$, then such quasigroups can easily be axiomatized using identities.
If $(X,*,/,\backslash)$ is a quasigroup, then define mappings $L_{a},R_{a}:X\rightarrow X$ by letting $L_{a}(x)=a*x,R_{a}(x)=x*a$.
Observation: Let $p$ be a prime. Let $(X,*,/,\backslash)$ be a quasigroup. The following are equivalent:

*

*if $a\neq b$, then all the cycles in the permutation $x\mapsto a\backslash(b*x)$ are finite of length $p$,


*$(X,*,/,\backslash)$ satisfies the identity
$(L_{x}L_{y}^{-1})^{p}(z)=z$.
The general case
The quasigroups associated with the Latin squares with one cycle type can be endowed with 5-ary algebraic operations that satisfy non-trivial identities.
Let $X$ be a finite set, and let $F_{X}$ denote the free group generated by $X$. Then we say than an element $h\in F_{X}$ is Brunnian if whenever $\phi:F_{X}\rightarrow G$ is a group homomorphism such that $\phi(x)=e$ for some $x\in X$, then $\phi(h)=e$. For example, if $[x,y]=xyx^{-1}y^{-1}$, and $X=\{x_{1},\dots,x_{n}\},|X|=n$, then
the iterated commutator $[x_{1},[x_{2},[\dots[x_{n-1},x_{n}]\dots]]]$ is always Brunnian.
Suppose that $(X,*,/,\setminus)$ is a finite quasigroup such that if
$a\neq b,c\neq d$, then the mappings $x\mapsto a\setminus(b*x),x\mapsto c\setminus(d*x)$ have the same cycle type.
Let $s,t:X^{5}\rightarrow X$ be 5-ary operations such that
$s(a,b,c,d,t(a,b,c,d,x))=t(a,b,c,d,s(a,b,c,d,x))=x$ whenever $a,b,c,d,x\in X$ and where if $a\neq b,c\neq d$, then
$c\setminus(d*x)=s(a,b,c,d,a\setminus (b*t(a,b,c,d,x))).$
If $a\in X$, then define permutations $L_{a},R_{a}:X\rightarrow X$ by letting
$L_{a}(x)=a*x,R_{a}(x)=x*a$.
Suppose now that $a,b,c,d\in X$. Then define a permutation $h_{a,b,c,d}$
$X$ by letting $h_{a,b,c,d}(x)=t(a,b,c,d,x).$
Let $w\in F=F_{l_{1},l_{2},l_{3},l_{4},r_{1},r_{2},r_{3},r_{4},q}$ be an element such that
i. if $\phi:F\rightarrow G$ is a group homomorphism with $\phi(l_{3}^{-1}l_{4})=\phi(q^{-1}l_{1}^{-1}l_{2}q)$, then $\phi(w)=e$,
ii. if $\phi:F\rightarrow G$ is a group homomorphism with $\phi(l_{1})=\phi(l_{2}),\phi(r_{1})=\phi(r_{2})$, then $\phi(w)=e$, and
iii. if $\phi:F\rightarrow G$ is a group homomorphism with $\phi(l_{3})=\phi(l_{4}),\phi(r_{3})=\phi(r_{4})$, then $\phi(w)=e$.
For example, every Brunnian element in $F$ satisfies the conditions i-iii.
Define a homomorphism $\psi_{a_{1},a_{2},a_{3},a_{4}}:F\rightarrow\text{Sym}(X)$ by letting
$\psi(q)=h_{a_{1},a_{2},a_{3},a_{4}}$ and $\psi(l_{k})=L_{a_{k}},\psi(r_{k})=R_{a_{k}}$ whenever $1\leq k\leq 4$. Then the algebraic structure $(X,*,/,\backslash,s,t)$ satisfies the identity
$\psi_{a_{1},a_{2},a_{3},a_{4}}(w)(x)=x$.
