A Latin square $\Lambda$ over an alphabet $A$ is a set of triples of elements of $A$ such that for every $\alpha,\beta\in A$, there is exactly one $\gamma\in A$ for which $(\alpha,\beta,\gamma)\in \Lambda$. A quasi-group $Q$ is a set with a binary operation $\cdot$ such that each of the equations $a\cdot x=b$ and $y\cdot a =b$ has exactly one solution $x,y\in Q$. Clearly, the set of triples $(\alpha,\beta,\gamma)$ of elements of $Q$ for which $\alpha\cdot\beta=\gamma$ forms a Latin square.
We can think of a Latin square $\Lambda$ over an alphabet $A$ of size $n$ to be an $n\times n$ array with rows and columns labelled by elements of $A$ and the $(\alpha,\beta)$ entry is $\gamma$ if $(\alpha,\beta,\gamma)\in \Lambda$.
Let $\Lambda, \Lambda'$ be Latin squares over an alphabet $A$. An isotopism is a triple $(\pi_1,\pi_2,\pi_3)$ of permutations of $A$such that for all $(\alpha,\beta,\gamma)\in \Lambda$, $(\alpha^{\pi_1},\beta^{\pi_2},\gamma^{\pi_3})\in \Lambda'$. Intuitively, an isotopism is a permutation of the labels of the rows, columns and entries. We say that $\Lambda$ and $\Lambda'$ are conjugate if we can obtain $\Lambda'$ from $\Lambda$ by swapping the elements $i$ and $j$ for every triple in $\Lambda$ for distinct $i,j\in \{1,2,3\}$. For example, if $i=1$ and $j=2$ this corresponds to swapping the labels of the rows and columns, so it would be a transposition, if Latin squares are viewed as matrices. A combination of isotopisms and conjugations is referred to as a parastrophism.
Now, it is a well known theorem by Albert that if $G$ and $H$ are groups and $\Lambda_G$, $\Lambda_H$ the corresponding Latin squares, then $G$ and $H$ are isomorphic if, and only if, $\Lambda_G$ and $\Lambda_H$ are related by some parastrophism.
My question is the following: are there known examples of non-isomorphic quasi-groups $Q$ and $R$ for which the corresponding Latin squares are related by some parastrophism?
If I understand correctly: we want a Latin square $Q$ for which some non-trivial conjugate $Q'$ is not isotopic to $Q$. And $R$ can be any isotope of $Q'$ (since $Q$ and $Q'$ are isotopic if and only if $Q$ and $R$ are isotopic).
Almost all Latin squares have no non-trivial autoparatopisms (McKay and Wanless 2005) and they exist for orders $\geq 7$ (and don't exist for orders $\leq 6$) (Phelps 1980). Thus, $Q$ and $R$ exist for orders $\geq 7$, and a random Latin square $Q$ and a random non-trivial conjugate $Q'$ with high probability has this property.
If you generate a random Latin square $Q$ of order $7$ and a random non-trivial conjugate $Q'$, you'll probably find one. As a concrete example: $$Q=\begin{array}{|ccccccc|} \hline 7 & 3 & 5 & 4 & 6 & 2 & 1 \\ 4 & 2 & 6 & 3 & 7 & 1 & 5 \\ 5 & 1 & 4 & 2 & 3 & 6 & 7 \\ 1 & 5 & 2 & 6 & 4 & 7 & 3 \\ 6 & 7 & 3 & 1 & 5 & 4 & 2 \\ 3 & 6 & 1 & 7 & 2 & 5 & 4 \\ 2 & 4 & 7 & 5 & 1 & 3 & 6 \\ \hline \end{array} $$
An easy way to prove it's not isotopic to its transpose $Q^T$ (which is one of its parastrophes) is to count the number of $2 \times 2$ subsquares intersecting each cell: $$ \begin{array}{|ccccccc|} \hline 0 & 0 & 0 & 1 & 2 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 2 & 1 & 0 & 1 \\ 2 & 0 & 1 & 2 & 1 & 0 & 2 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ \bf 3 & \bf 2 & \bf 0 & \bf 2 & \bf 2 & \bf 0 & \bf 3 \\ \hline \end{array} $$ We find the Latin square has a row containing 2 cells which intersect three $2 \times 2$ subsquares. If it were isotopic to its transpose, then there would be a column with this property, but there is no such column. Thus it's not isotopic to its transpose.