If $A^3=A$, then $A$ satisfies the polynomial $t^3-t = t(t^2-1)=t(t-1)(t+1)$.
Since you say you know about the minimal polynomial
The minimal polynomial divides every polynomial that the matrix satisfies, so the minimal polynomial divides $t(t-1)(t+1)$. In particular, since it is squarefree, $A$ must be diagonalizable.
If the matrix has a single eigenvalue, then it is a scalar multiple of the identity; the three possibilities are $0$, $I$, and $-I$.
If the matrix has distinct eigenvalues, then the eigenvalues are either $0$ and $1$, $0$ and $-1$, or $1$ and $-1$. In any of these cases, there are infinitely many distinct matrices that satisfy these conditions. For example, for eigenvalues $0$ and $1$, any projection onto a 1-dimensional subspace will do; there are infinitely many different projections onto, say, the subspace $\{(x,0)\mid x\in\mathbb{C}\}$, one for every possible complement. That already solves the problem.
If you don't remember about the minimal polynomial, you can note that any eigenvalue $\lambda$ must be a root of the polynomial $t^3-t$ (more generally, if $A$ satisfies $f(t)$, then every eigenvalue of $A$ must be a root of $f(t)$). That leads you to the fact that $A$ has eigenvalues $0$, $1$, and $-1$, and from there you can obtain infinitely many different matrices as noted above.