imc

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: $384(3)=\boxed{1152}$