It's Not A Magic Square

The only important things to consider are that even numbers (2, 4, 6, 8) may not be adjacent to each other, and similarly for the multiples of 3 (3, 6, 9). There is no other case where two numbers in the range 1-9 are not relatively prime.

Thus we can now start a massive case checking.

First, color the board with black and white like a chessboard, so the corners are black. All the even numbers must go in a single color. This can be verified by case checking again: if a white square is occupied (by an even number), there are three black squares that cannot be occupied; if two white squares are occupied, there are at least four black squares that cannot be occupied; if three white squares are occupied, all black squares cannot be occupied.

Case 1 : All even numbers go to white squares.

This case is much simpler. There are $4!$ ways to permute the even numbers. No matter how they are permuted, the three squares next to the 6 can only have the numbers 1, 5, 7 (the 3 and 9 cannot be adjacent to them); there are $3!$ ways to permute them. The 3 and 9 are for the other two squares, with $2!$ ways to permute. In total, there are $4! \cdot 3! \cdot 2! = 288$ ways here.

Case 2 : All even numbers go to black squares.

This case has two more subcases, depending on whether the center square is an even number or not.

Case 2.1 : The center square is not an even number.

There are $4!$ ways to permute the even numbers to the corners. In any case, the two squares next to the 6 cannot be filled with 3 or 9 (adjacent and not coprime to 6), and the center cannot be either (adjacent to all the remaining empty squares, and hence the other of 3 and 9, and not coprime with that other number). There are thus $2!$ ways to permute the 3 and 9, and $3!$ ways for the 1, 5, 7 in the remaining spots. In total, this is $4! \cdot 2! \cdot 3! = 288$ ways.

(Diagram: Squares marked x below cannot be 3 or 9.)

1
2
3

6 x E
x x O
E O E

Case 2.2 : The center square is an even number.

The center square cannot be 6, as it's adjacent to four of the five remaining empty squares. So the 6 is on a corner. After that, this is, again, divided into two cases, depending on the location of 6 and the empty corner (the corner with an odd number):

1
2
3

6 O E | E O E
O E O | O E O
E O O | 6 O O

Case 2.2.1 : 6 is across the empty corner.

There are $4$ ways to select the position of the empty corner; this uniquely determines the position of the 6. There are $3!$ ways to permute the even numbers. Now, similar to above, the positions of 3 and 9 are also fixed; there are $2!$ ways to put them, and $3!$ for the 1, 5, 7, for a total of $4 \cdot 3! \cdot 2! \cdot 3! = 288$ .

3 and 9 cannot be on the squares marked x below.

1
2
3

6 x E
x E O
E O x

Case 2.2.2 : 6 is not across the empty corner.

There are $4$ ways to select the position of the empty corner, and $2$ ways to select the position of the 6. There are $3!$ ways to select the other even numbers. This time, there are two possibilities for the positions of 3 and 9. The top O below is 3 or 9, and one of the remaining two O's is also 3 or 9, but either one works. That's a total of $2$ ways to choose the positions, and $2!$ ways to permute 3 and 9. There are $3!$ more ways for the 1, 5, 7, for a total of $4 \cdot 2 \cdot 3! \cdot 2 \cdot 2! \cdot 3! = 1152$ ways.

1
2
3

E O E
x E O
6 x O

Totaling everything, we have $288 + 288 + 288 + 1152 = \boxed{2016}$ ways.