Answer: $a = 0$ or $a = -1$.

By condition, there is $c$ such that $f(c) = a$. Substituting $c$ for $x$ in the equality
$$f(f(x)) = x f(x) - a x, \quad (1)$$
we obtain $f(a) = f(f(c)) = c f(c) - a c = c a - a c = 0$.

Substituting $a$ for $x$ in (1), we obtain
$$f(0) = f(f(a)) = a f(a) - a^2 = -a^2 \quad (2)$$

Substituting $0$ for $x$ in (1), we obtain $f(-a^2) = f(f(0)) = 0 \cdot f(0) - a \cdot 0 = 0$.

Substituting $-a^2$ for $x$ in (1), we obtain
$$f(0) = f(f(-a^2)) = -a^2 f(-a^2) - a \cdot (-a^2) = a^3. \quad (3)$$

From (2) and (3) it follows that $-a^2 = a^3$, hence either $a = 0$ or $a = -1$.

If $a = -1$, then it is easy to see that the function
$$f(x) = \begin{cases} -1 & \text{for } x \ne -1, \\ 0 & \text{for } x = -1 \end{cases}$$
satisfies the problem condition.

If $a = 0$, then it is easy to see that the function
$$f(x) = \begin{cases} 0 & \text{for } x \ne 1, \\ 1 & \text{for } x = 1 \end{cases}$$
satisfies the problem condition.