jmc

The diagonals of a rectangle are of the same length and bisect each other. The diagonal determined by the two given points has its midpoint at $(0,0),$ and it has length $$\sqrt{(4-(-4){)}^2+(3-(-3){)}^2}=\sqrt{8^2+6^2}=10.$$Thus, the other diagonal must have its midpoint at $(0,0)$ and also have length $10.$ This means that the other two vertices of the rectangle must lie on the circle $x^2+y^2=5^2.$ The number of lattice points (points with integer coordinates) on this circle is $12$: we have $$(\pm 4,\pm 3),(\pm 3,\pm 4),(\pm 5,0),(0,\pm 5).$$These points pair up to give $6$ possible diagonals, of which $1$ is the given diagonal. Therefore, there are $6-1=5$ choices for the other diagonal, which uniquely determines a rectangle. Thus there are $\boxed{5}$ possible rectangles.