Solution:
$-1$ and $2$

Let $f^{(n)}(x) = (\underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}})(x)$. For allowed values of $x$, note that $f^{(n)}(x)$ is of the form
$$f^{(n)}(x) = \frac{a_n x + b_n}{c_n x + d_n}$$
where $a_n, b_n, c_n, d_n \in \mathbb{Z}$ for all integers $n \geq 1$. The equation
$$\frac{a_n x + b_n}{c_n x + d_n} = x \quad \text{or} \quad c_n x^2 + (d_n - a_n)x - b_n = 0$$
has at most two real roots. Since $f(-1) = -1$ and $f(2) = 2$, it follows that $f^{(n)}(-1) = -1$ and $f^{(n)}(2) = 2$ for all $n \geq 1$. Thus, the roots of $f^{(10)}(x) = x$ are $-1$ and $2$.