, 2016

First notice that choosing two of the numbers in adjacent squares completely determines the remaining numbers in that row or column.
If we can inscribe the numbers as required then any permutation of the labels $1$, $2$ and $3$ will also give us a correct solution. We can put the numbers $1$, $2$ and $3$ in $6$ different orders, namely $(1, 2, 3)$, $(1, 3, 2)$, $(2, 1, 3)$, $(2, 3, 1)$, $(3, 1, 2)$ and $(3, 2, 1)$.
So, we have to count the arrays which contain $1$, $2$, $3$, in this order, in the first three squares of the first row, and multiply the result by $6$.
Placing $1$, $2$ and $3$ in the first three squares of the first row gives us two possibilities for the first square of the second row, $2$ or $3$. In both cases the remainder of the upper left $3 \times 3$ square is then uniquely determined and is

or
Using the same argument we can show that the remainder of the array is now also uniquely determined. In the first case each of the right-up diagonals will contain the same number with each of the number repeating on every third diagonal. In the second case each of the right-down diagonals will contain the same number with each number repeating on every third diagonal. In both cases the array will be filled correctly since it will not be possible for the tile to cover the same number twice.
We conclude that there are two ways to fill the array in this way. Allowing for permutations, there are $2 \cdot 6 = 12$ solutions.