Solution:
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 $3 \cdot 12 = 36$.