The regular octagon $ABCDEFGH$ has its center at $J$.  Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal.  In how many ways can this be done?
$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576  \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$
First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:

We also notice that $A+E = B+F = C+G = D+H$.  
WLOG, assume that $J = 1$.  Thus the pairs of vertices must be $9$ and $2$, $8$ and $3$, $7$ and $4$, and $6$ and $5$.  There are $4! = 24$ ways to assign these to the vertices.  Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$).  
Thus, there are $16(24) = 384$ ways for each possible J value.  There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{\textbf{(C) }1152}$