Denote the number of cell $(i, j)$ in the square by $a_{ij}$. For $i, j, r, k \in \{1, 2, \dots, n\}$, with $i \neq j$ and $r \neq k$, we define the operation $P_{ijrk}$ in this way: Add $1$ to all the cells in the union of row and column of cell $(i, j)$, add $1$ to all the cells in the union of row and column of cell $(r, k)$, add $-1$ to all the cells in the union of row and column of cell $(i, k)$, and finally add $-1$ to all the cells in the union of row and column of cell $(r, j)$.
After this operation, $a_{ir}, a_{jk}$ will increase by $1$, $a_{ik}, a_{rk}$ will decrease by $1$ and all the other cells remain the same.

For two squares $A$ and $B$ we define
$$d(A, B) = \sum_{1 \le i, j \le n} |a_{ij} - b_{ij}|.$$
We want to use the above operation to replace the Latin square $A$ by some other square $A'$ such that $d(A', B) < d(A, B)$ if we can do this by repeating this operation we can transform $A$ to $B$.
Because the sum of the numbers in each row and column in $A$ and $B$ are the same, there should be two integer numbers $1 \le i, j \le n$ so that $a_{ij} > b_{ij}$ and two integer numbers $1 \le k, l \le n$ so that $a_{ik} < b_{ik}, a_{lj} < b_{lj}$. Now if we do the operation $P_{iljk}$ on $A$ to obtain $A'$ then $d(A', B) < d(A, B)$ and the sum of numbers in each row and column of $A'$ does not change. So we can reach $B$ by repeating this process. ■