A Latin square $A$ is called self-orthogonal if $A$ and $A^{T}$ are orthogonal Latin squares.
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 $a_{ij}=2i-j\in \mathbb{Z}_v$. Prove that this forms a self-orthogonal Latin square if the $\gcd(v,6)=1$. I have just tried to see that this works with small orders like $v=5$: $$\boldsymbol{A}=\begin{pmatrix} 0 & 4 & 3 & 2 & 1\\ 2 & 1 & 0 & 4 & 3\\ 4 & 3 & 2 & 1 & 0\\ 1 & 0 & 4 & 3 & 2\\ 3 & 2 & 1 & 0 & 4\\ \end{pmatrix}\implies\boldsymbol{A}^{T}=\begin{pmatrix} 0 & 2 & 4 & 1 & 3\\ 4 & 1 & 3 & 0 & 2\\ 3 & 0 & 2 & 4 & 1\\ 2 & 4 & 1 & 3 & 0\\ 1 & 3 & 0 & 2 & 4\\ \end{pmatrix}$$ So it seems to work and I could probably sketch a general construction, but I don't know how the assumption $\gcd(v,6)=1$ makes this all work. Can someone explain?
Let's assume $\gcd(v, 6)=1$ and work in the ring $\Bbb{Z}_v$ except where otherwise stated (so I can write $i = i'$ rather than $v \mid i - i'$ or $i \equiv i' \,(\mathrm{mod}\,v)$).
We have $a_{ij} = a_{ij'}$ iff $2i -j = 2i - j'$ iff $j = j'$. Also we have $a_{ij} = a_{i'j}$ iff $2i - j = 2i' - j$ iff $2(i - i') = 0$ iff $i = i'$, since $2$ is invertible in $\Bbb{Z}_v$ as $\gcd(v, 6) \neq 1$. Hence $A = (a_{ij})_{i, j \in \Bbb{Z}{_v}}$ is a Latin square.
$A$ is self-orthogonal if the pairs $(A_{ij}, A^T_{ij}) = (a_{ij}, a_{ji})$ are all distinct. but $(a_{ij}, a_{ji}) = (a_{i'j'}, a_{j'i'})$ iff the following equations hold: $$ \begin{align*} 2i - j &= 2i' - j' \\ 2j - i &= 2j' - i' \end{align*} $$ Since $2$ is invertible as $\gcd(v, 6) = 1$, we can rewrite these equations as: $$ \begin{align*} i - i' &= \frac{1}{2}(j - j') \\ j - j' &= \frac{1}{2}(i - i') \end{align*} $$ whence $$ \begin{align*} i - i' &= \frac{1}{4}(i - i') \\ j - j' &= \frac{1}{4}(j - j') \end{align*} $$
which gives us $$ \begin{align*} 4(i - i') &= i - i' \\ 4(j - j') &= j - j' \end{align*} $$
but if $4x = x$ in $\Bbb{Z}_v$, we must have that $v \mid 3x$ in $\Bbb{Z}$ implying that $x =0$ in $\Bbb{Z}_v$, since $v$ and $3$ are coprime by our assumption that $\gcd(v, 6) = 1$. Hence $i = i'$ and $j = j'$ and $A$ is indeed self-orthogonal.