Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that for every prime number $p$ and natural number $x$,
$$\{ x,f(x),\cdots f^{p-1}(x) \} $$
is a complete residue system modulo $p$. With $f^{k+1}(x)=f(f^k(x))$ for every natural number $k$ and $f^1(x)=f(x)$.

[i]Proposed by IndoMathXdZ[/i]
1. **Understanding the Problem:**
   We need to find all functions \( f: \mathbb{N} \rightarrow \mathbb{N} \) such that for every prime number \( p \) and natural number \( x \), the set \( \{ x, f(x), f^2(x), \ldots, f^{p-1}(x) \} \) forms a complete residue system modulo \( p \). This means that the set contains \( p \) distinct elements, each of which is congruent to a different integer modulo \( p \).

2. **Initial Observation:**
   If \( |f(x) - x| \geq 2 \), consider a prime \( p \) dividing \( |f(x) - x| \). This would imply that \( f(x) \equiv x \pmod{p} \), which contradicts the requirement that \( \{ x, f(x), \ldots, f^{p-1}(x) \} \) forms a complete residue system modulo \( p \). Therefore, \( |f(x) - x| \) must be less than 2.

3. **Possible Values for \( f(x) \):**
   Since \( |f(x) - x| < 2 \), the only possibilities are \( f(x) = x + 1 \) or \( f(x) = x - 1 \).

4. **Eliminating \( f(x) = x - 1 \):**
   If \( f(x) = x - 1 \), then \( f^2(x) = x - 2 \), and so on. For a prime \( p \), the set \( \{ x, x-1, x-2, \ldots, x-(p-1) \} \) would not form a complete residue system modulo \( p \) because it would not cover all residues \( 0, 1, 2, \ldots, p-1 \). Hence, \( f(x) = x - 1 \) is not a valid solution.

5. **Verifying \( f(x) = x + 1 \):**
   If \( f(x) = x + 1 \), then \( f^2(x) = x + 2 \), and so on. For a prime \( p \), the set \( \{ x, x+1, x+2, \ldots, x+(p-1) \} \) modulo \( p \) will indeed form a complete residue system modulo \( p \) because it will cover all residues \( 0, 1, 2, \ldots, p-1 \).

6. **Inductive Step:**
   Assume \( f(n) = n + 1 \). We need to show that \( f(n+1) = n + 2 \). If \( f(n+1) = n \), then \( f(f(n)) = n \), which contradicts the assumption that \( f(n) = n + 1 \). Therefore, \( f(n+1) = n + 2 \).

7. **Conclusion:**
   The only function that satisfies the given condition is \( f(x) = x + 1 \) for all \( x \in \mathbb{N} \).

The final answer is \( \boxed{ f(x) = x + 1 } \).