Find all real numbers $a, b, c$ that satisfy
$$ 2a - b =a^2b, \qquad 2b-c = b^2 c, \qquad 2c-a= c^2 a.$$
To solve the system of equations:
\[ 2a - b = a^2b, \]
\[ 2b - c = b^2c, \]
\[ 2c - a = c^2a, \]
we will first analyze the function \( f(x) = \frac{2x}{1 + x^2} \).

1. **Fixed Points:**
   We start by finding the fixed points of \( f(x) \). A fixed point \( x \) satisfies:
   \[ f(x) = x \implies \frac{2x}{1 + x^2} = x. \]
   Solving this equation:
   \[ \frac{2x}{1 + x^2} = x \implies 2x = x(1 + x^2) \implies 2x = x + x^3 \implies x^3 - x = 0 \implies x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0. \]
   Therefore, the fixed points are:
   \[ x \in \{-1, 0, 1\}. \]

2. **Three-Cycle Points:**
   Next, we consider the three-cycle points. We need to find \( a, b, c \) such that:
   \[ a = f(f(f(a))). \]
   Using the trigonometric identity for tangent, we know:
   \[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}. \]
   Therefore, if \( a = i \tan \theta \), then:
   \[ f(i \tan \theta) = i \tan 2\theta. \]
   For a three-cycle, we need:
   \[ \tan 8\theta = \tan \theta \implies 8\theta = \theta + k\pi \implies \theta = \frac{k\pi}{7} \text{ for } k \in \mathbb{Z}. \]

3. **Real and Complex Solutions:**
   From the above analysis, we can derive the following solutions:
   - **Real Solutions:**
     \[ (-1, -1, -1), \quad (0, 0, 0), \quad (1, 1, 1). \]
   - **Complex Solutions:**
     \[ \left(i \tan \frac{\pi}{7}, i \tan \frac{2\pi}{7}, i \tan \frac{4\pi}{7}\right), \]
     \[ \left(i \tan \frac{2\pi}{7}, i \tan \frac{4\pi}{7}, i \tan \frac{\pi}{7}\right), \]
     \[ \left(i \tan \frac{4\pi}{7}, i \tan \frac{\pi}{7}, i \tan \frac{2\pi}{7}\right). \]

Since the problem asks for real numbers \( a, b, c \), we conclude that the only real solutions are:
\[ (-1, -1, -1), \quad (0, 0, 0), \quad (1, 1, 1). \]

The final answer is \( \boxed{ (-1, -1, -1), (0, 0, 0), (1, 1, 1) } \).