Note that 6 can only be adjacent to 1, 5, and 7, so there are $\binom{3}{2}=3$ ways to pick its neighbors. Since each of 1, 5, and 7 is relatively prime to every number in $\{1,2,3,4,5,6,7,8\}$ but itself (and hence can have arbitrary neighbors), without loss of generality suppose we have picked 1 and 5 as neighbors of 6. Observe that fixing the positions of 1, 5, and 6 eliminates the indistinguishability of rotations and reflections. Now, we have to consecutively arrange $\{2,3,4,7,8\}$ so that no two of 2, 4, and 8 are adjacent. There are $3!\cdot 2! = 12$ ways of doing so, so the final answer is $3 \cdot 12=36$