Find all three real numbers $(x, y, z)$ satisfying the system of equations $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x}{z}+\frac{z}{y}+\frac{y}{x}$$ $$x^2 + y^2 + z^2 = xy + yz + zx + 4$$
1. **Symmetry and Simplification:**
   The given system of equations is:
   \[
   \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \frac{x}{z} + \frac{z}{y} + \frac{y}{x}
   \]
   \[
   x^2 + y^2 + z^2 = xy + yz + zx + 4
   \]
   Since the expressions are symmetric, if \((x, y, z)\) is a solution, then any permutation of it is also a solution.

2. **Analyzing the First Equation:**
   Rearrange the first equation:
   \[
   \frac{x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{x}{z} - \frac{z}{y} - \frac{y}{x} = 0
   \]
   Combine like terms:
   \[
   \left(\frac{x}{y} - \frac{y}{x}\right) + \left(\frac{y}{z} - \frac{z}{y}\right) + \left(\frac{z}{x} - \frac{x}{z}\right) = 0
   \]
   Factor each term:
   \[
   \frac{x^2 - y^2}{xy} + \frac{y^2 - z^2}{yz} + \frac{z^2 - x^2}{zx} = 0
   \]
   Simplify:
   \[
   \frac{(x-y)(x+y)}{xy} + \frac{(y-z)(y+z)}{yz} + \frac{(z-x)(z+x)}{zx} = 0
   \]
   Multiply through by \(xyz\):
   \[
   (x-y)(x+y)z + (y-z)(y+z)x + (z-x)(z+x)y = 0
   \]
   This implies:
   \[
   (x-y)(y-z)(z-x) = 0
   \]
   Therefore, at least one of \(x = y\), \(y = z\), or \(z = x\) must be true.

3. **Case Analysis:**
   Without loss of generality (WLOG), assume \(y = z\).

4. **Substitute \(y = z\) into the Second Equation:**
   The second equation becomes:
   \[
   x^2 + y^2 + y^2 = xy + yy + yx + 4
   \]
   Simplify:
   \[
   x^2 + 2y^2 = 2xy + y^2 + 4
   \]
   Rearrange:
   \[
   x^2 + y^2 - 2xy = 4
   \]
   Factor:
   \[
   (x - y)^2 = 4
   \]
   Therefore:
   \[
   x - y = 2 \quad \text{or} \quad x - y = -2
   \]

5. **Solutions:**
   - If \(x - y = 2\), then \(x = y + 2\).
   - If \(x - y = -2\), then \(x = y - 2\).

   Thus, the solutions are:
   \[
   (x, y, z) = (y+2, y, y) \quad \text{or} \quad (y-2, y, y)
   \]
   Considering permutations, the complete set of solutions is:
   \[
   (x, x-2, x-2) \quad \text{and} \quad (x, x+2, x+2)
   \]
   where \(x \in \mathbb{R}\).

The final answer is \( \boxed{ (x, x-2, x-2) } \) and \((x, x+2, x+2)\) for \(x \in \mathbb{R}\).