3. (10 points) In a $3 \times 3$ grid (each cell is a $1 \times 1$ square), place two identical pieces, with at most one piece per cell, there are $\qquad$ different ways to place them. (If two placements can be made to coincide by rotation, they are considered the same placement).
【Analysis】We can discuss the problem by dividing it into different cases. The four vertices have the same position value, the square right in the middle has one position value, and the remaining four squares have the same position value. Therefore, we can calculate the different placement methods for each of the three cases.

【Solution】Solution: According to the analysis, we divide it into three cases:
(1) When a chess piece is placed at the center, i.e., at $E$, and the other chess piece is placed at any position on the periphery, excluding symmetry factors, there are 2 different placement methods, i.e., $A E$, $B E$;
(2) When neither of the two chess pieces is at the center $E$, and one of them is at a vertex, there are 4 different placement methods, i.e., $A B$, $A F$, $A H$, $A D$;
(3) When both chess pieces are at the vertices, there are 2 different placement methods, i.e., $A C$, $A I$;
(4) When both chess pieces are in the 4 squares other than the vertices and the center, there are 2 different placement methods, i.e., $B D$, $B H$. In summary, there are: $2+4+2+2=10$ different placement methods.