Solution:

2016

The numbers can be separated into four sets. Numbers in the set $A=\{1,5,7\}$ can be placed next to anything. The next two sets are $B=\{2,4,8\}$ and $C=\{3,9\}$. The number $6$, which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be interchanged without violating the condition, so without loss of generality, we can pretend we have three $1$'s, three $2$'s, two $3$'s, and one $6$, as long as we multiply our answer by $3!3!2!$ at the end. The available arrangements are, grouped by the position of the $6$, are:

When $6$ is in contact with three numbers:

When $6$ is in contact with two numbers:

The next two can be flipped diagonally to create different arrangements:

Those seven arrangements can be rotated $90$, $180$, and $270$ degrees about the center to generate a total of $28$ arrangements. $28 \cdot 3!3!2! = 2016$.