For $1 \leq i, j \leq n$, choose $a_{i, j}$ equal to the remainder of $3i + j$ when divided by $n$. Clearly, $0 \leq a_{i, j} \leq n-1$, for $1 \leq i, j \leq n$.

If we fix a row $1 \leq i \leq n$, $a_{i, j_1} = a_{i, j_2}$, for $1 \leq j_1, j_2 \leq n$, implies $3i + j_1 \equiv 3i + j_2 \pmod{n}$, that is $j_1 \equiv j_2 \pmod{n}$ and therefore $j_1 = j_2$.

Hence, $a_{i, 1}, \ldots, a_{i, n}$ is a permutation of $0, 1, \ldots, n-1$.

If we fix a column $1 \leq j \leq n$, $a_{i_1, j} = a_{i_2, j}$, for $1 \leq i_1, i_2 \leq n$, implies $3i_1 + j \equiv 3i_2 + j \pmod{n}$, that is $3i_1 \equiv 3i_2 \pmod{n}$. But $n$ is not divisible by $3$. Therefore $i_1 = i_2$. This proves that $a_{1, j}, \ldots, a_{n, j}$ is a permutation of $0, 1, \ldots, n-1$.

Now, let $1 \leq i < j \leq n$ and $1 \leq i' < j' \leq n$ such that $(a_{i, j}, a_{j, i}) = (a_{i', j'}, a_{j', i'})$ this implies that $3i + j \equiv 3i' + j' \pmod{n}$ and $3j + i \equiv 3j' + i' \pmod{n}$. This is equivalent to $8i \equiv 8i' \pmod{n}$ and $8j \equiv 8j' \pmod{n}$. But $n$ is an odd number. Therefore, $(i, j) = (i', j')$. Hence, the board is totally non-symmetric.