For which quadratic polynomials $f(x)$ does there exist a quadratic polynomial $g(x)$ such that the equations $g(f(x)) = 0$ and $f(x)g(x) = 0$ have the same roots, which are mutually distinct and form an arithmetic progression?
1. **Identify the roots of \( f(x) \) and \( g(x) \):**
   Let \( f(x) \) be a quadratic polynomial with roots \( a \) and \( b \), and let \( g(x) \) be a quadratic polynomial with roots \( c \) and \( d \).

2. **Equating the roots of \( f(x)g(x) \) and \( g(f(x)) \):**
   Since \( f(x)g(x) = 0 \) and \( g(f(x)) = 0 \) have the same roots, we have:
   \[
   g(f(a)) = g(f(b)) = g(0) = 0
   \]
   This implies that \( 0 \) is a root of \( g(x) \). Without loss of generality, let \( c = 0 \).

3. **Determine the form of \( g(x) \):**
   Since \( 0 \) is a root of \( g(x) \), we can write \( g(x) \) as:
   \[
   g(x) = x(x - d)
   \]
   where \( d \) is the other root of \( g(x) \).

4. **Evaluate \( g(f(0)) \) and \( g(f(d)) \):**
   Given that the roots are distinct and form an arithmetic progression, we have:
   \[
   g(f(0)) = g(f(d)) = 0
   \]
   This implies:
   \[
   f(0) = 0 \quad \text{or} \quad f(0) = d
   \]
   and
   \[
   f(d) = 0 \quad \text{or} \quad f(d) = d
   \]

5. **Determine the roots of \( f(x) \):**
   Since the roots form an arithmetic progression, let the roots of \( f(x) \) be \( a \) and \( 2a \). Then, \( d = 3a \).

6. **Formulate \( f(x) \):**
   Given \( f(0) = 0 \) and \( f(3a) = 3a \), and knowing that \( f(a) = f(2a) = 0 \), we can write:
   \[
   f(x) = k(x - a)(x - 2a)
   \]
   To find \( k \), use the condition \( f(3a) = 3a \):
   \[
   f(3a) = k(3a - a)(3a - 2a) = k(2a)(a) = 2ka^2
   \]
   Setting \( f(3a) = 3a \), we get:
   \[
   2ka^2 = 3a \implies k = \frac{3}{2a}
   \]
   Therefore, the quadratic polynomial \( f(x) \) is:
   \[
   f(x) = \frac{3}{2a}(x - a)(x - 2a)
   \]

The final answer is \( \boxed{ f(x) = \frac{3}{2a}(x - a)(x - 2a) } \)