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 equal.  In how many ways can this be done?

$\textbf{(A) }384\qquad
\textbf{(B) }576\qquad
\textbf{(C) }1152\qquad
\textbf{(D) }1680\qquad
\textbf{(E) }3546\qquad$

[asy]
size(175);
defaultpen(linewidth(0.8));
path octagon;
string labels[]={"A","B","C","D","E","F","G","H","I"};
for(int i=0;i<=7;i=i+1)
{
pair vertex=dir(135-45/2-45*i);
octagon=octagon--vertex;
label("$"+labels[i]+"$",vertex,dir(origin--vertex));
}
draw(octagon--cycle);
dot(origin);
label("$J$",origin,dir(0));
[/asy]
1. **Define Variables and Equations:**
   Let the numbers assigned to vertices \(A, B, C, D, E, F, G, H\) be \(a, b, c, d, e, f, g, h\) respectively, and the number assigned to the center \(J\) be \(j\). We need the sums of the numbers on the lines \(AJE, BJF, CJG, DJH\) to be equal. This gives us the equations:
   \[
   a + j + e = b + j + f = c + j + g = d + j + h
   \]
   Simplifying, we get:
   \[
   a + e = b + f = c + g = d + h = S
   \]

2. **Sum of All Digits:**
   The digits \(1\) through \(9\) sum to:
   \[
   1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
   \]
   Including the center \(J\), the total sum is:
   \[
   a + b + c + d + e + f + g + h + j = 45
   \]
   Since \(a + e + b + f + c + g + d + h = 45 - j\), and each pair sums to \(S\), we have:
   \[
   4S = 45 - j
   \]
   Solving for \(j\):
   \[
   j = 45 - 4S
   \]

3. **Possible Values for \(j\):**
   Since \(j\) must be a digit from \(1\) to \(9\), we solve:
   \[
   1 \leq 45 - 4S \leq 9
   \]
   This simplifies to:
   \[
   36 \leq 4S \leq 44
   \]
   Dividing by 4:
   \[
   9 \leq S \leq 11
   \]
   Therefore, \(S\) can be \(9, 10,\) or \(11\).

4. **Case Analysis:**
   - **Case 1: \(S = 11\)**
     \[
     j = 45 - 4 \times 11 = 1
     \]
     - **Case 2: \(S = 10\)**
     \[
     j = 45 - 4 \times 10 = 5
     \]
     - **Case 3: \(S = 9\)**
     \[
     j = 45 - 4 \times 9 = 9
     \]

5. **Counting Arrangements:**
   For each case, we need to count the number of ways to assign the digits to the pairs \((a, e), (b, f), (c, g), (d, h)\) such that their sums are \(S\).
   - There are \(4!\) ways to assign the pairs.
   - Each pair can be arranged in \(2\) ways.
   - Therefore, for each case, the number of ways is:
     \[
     4! \times 2^4 = 24 \times 16 = 384
     \]

6. **Total Number of Ways:**
   Since there are 3 cases, the total number of ways is:
   \[
   3 \times 384 = 1152
   \]

Conclusion:
\[
\boxed{1152}
\]