Solution:

Note that $1$ and $9$ must be in the top left and bottom right corners respectively. We must also have that $2$ and $8$ are next to these squares.

We can then case on the center square. Note that the only possible values are $4,$$5,$ or $6.$

Case 1: the center is $4$

The $3$ is necessarily next to the $1,$ since there is no other option that is less than $4.$

Any number can be in the square next to the $8,$ but the other two squares are then fixed. There are $$2 \cdot 2 \cdot 3 = 12$$ cases (two places for the $2,$ two places for the $8,$ and three choices for the square adjacent to $8$).

Case 2: the center is $5$

We can case on the position of the $3.$ If the $3$ is in the top right square, the $4$ is necessarily next to the $1.$

If the $8$ is above the $9,$ then the other two squares are fixed. If it is to the left of the $9,$ the other two squares can be filled arbitrarily.

Now consider when the $3$ is below the $1.$ There are two spots for the $8,$ and the square next to the $8$ can be any number.

The other two squares are then fixed. This means that this case has a total of $$2(1 + 2 + 2 \cdot 3) = 18.$$ We multiply by two since the $2$ can be either to the right of or below the $1.$

Case 3: the center is $6$

This is similar to case $1$ since the $7$ is fixed instead of the $3.$

The total number of arrangements is then $$12 + 18 + 12 = 42.$$

Thus, D is the correct answer.