I have been unable to find an elegant method of constructing self orthogonal Latin squares. However, I came across this question: construct a self orthogonal Latin square of order 5 using the fact that the set of elements on the main diagonal of a self orthogonal Latin from a transversal. This seems to imply that you can use this fact alone to construct self orthogonal Latin squares or at-least ones of order five.
I have been unable to figure out how I can do this question and would appreciate any help.
Use the elements of $\;\mathbb{Z}_v$ as the names of the rows and columns of your Latin square. Let $\boldsymbol{A}=(a_{ij})$ such that $a_{ij}=2i-j\in \mathbb{Z}_v$. This forms a self-orthogonal Latin square always when the $\gcd(v,6)=1$, so it works for the order $5$ case as well.