Determine all integer solutions to the system of equations:
\begin{align*}
xy + yz + zx &= -4 \\
x^2 + y^2 + z^2 &= 24 \\
x^{3} + y^3 + z^3 + 3xyz &= 16
\end{align*}
1. **Given System of Equations:**
   \[
   \begin{cases}
   xy + yz + zx = -4 \\
   x^2 + y^2 + z^2 = 24 \\
   x^3 + y^3 + z^3 + 3xyz = 16
   \end{cases}
   \]

2. **Sum of Cubes Identity:**
   Recall the identity for the sum of cubes:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)
   \]
   Using this identity, we can rewrite the third equation as:
   \[
   x^3 + y^3 + z^3 + 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) + 6xyz
   \]

3. **Substitute Known Values:**
   Given \(x^2 + y^2 + z^2 = 24\) and \(xy + yz + zx = -4\), we substitute these into the identity:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x+y+z)(24 - (-4)) = (x+y+z)(28)
   \]
   Since \(x^3 + y^3 + z^3 + 3xyz = 16\), we have:
   \[
   (x+y+z)(28) + 6xyz = 16
   \]

4. **Determine \(x+y+z\):**
   Let \(s = x + y + z\) and \(p = xyz\). Then:
   \[
   28s + 6p = 16
   \]
   Solving for \(p\):
   \[
   6p = 16 - 28s \implies p = \frac{16 - 28s}{6} = \frac{8 - 14s}{3}
   \]
   Since \(p\) must be an integer, \(8 - 14s\) must be divisible by 3. Testing possible integer values for \(s\), we find:
   \[
   8 - 14s \equiv 0 \pmod{3} \implies 2 - 2s \equiv 0 \pmod{3} \implies s \equiv 1 \pmod{3}
   \]
   Thus, \(s\) can be \(1\) or \(-2\) (since \(s\) must be an integer).

5. **Check Possible Values of \(s\):**
   - If \(s = 4\):
     \[
     p = \frac{8 - 14 \cdot 4}{3} = \frac{8 - 56}{3} = \frac{-48}{3} = -16
     \]
   - If \(s = -4\):
     \[
     p = \frac{8 - 14 \cdot (-4)}{3} = \frac{8 + 56}{3} = \frac{64}{3} \quad \text{(not an integer)}
     \]

6. **Determine \(x, y, z\):**
   Given \(s = 4\) and \(p = -16\), we need to find \(x, y, z\) such that:
   \[
   x + y + z = 4, \quad xy + yz + zx = -4, \quad xyz = -16
   \]
   The polynomial whose roots are \(x, y, z\) is:
   \[
   t^3 - 4t^2 - 4t + 16 = 0
   \]
   Testing possible integer roots, we find:
   \[
   t = 2 \quad \text{is a root.}
   \]
   Using synthetic division:
   \[
   t^3 - 4t^2 - 4t + 16 = (t - 2)(t^2 - 2t - 8)
   \]
   Solving \(t^2 - 2t - 8 = 0\):
   \[
   t = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2} = 4 \text{ or } -2
   \]
   Thus, the solutions are \(x, y, z = 2, -2, 4\) and their permutations.

The final answer is \(\boxed{(2, -2, 4)}\) and their permutations.