Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}__0 \to \mathbb{N}_0$ such that:
\[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0  \]
where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?
To determine if there exists a function \( f: \mathbb{N}_0 \to \mathbb{N}_0 \) such that \( f^{2003}(n) = 5n \) for all \( n \in \mathbb{N}_0 \), we need to construct such a function and verify its properties.

1. **Define the sets \( S_i \):**
   - Group the natural numbers indivisible by 5 into groups of size \( k \). For example, if \( k = 3 \), the groups are \( (1, 2, 3) \), \( (4, 6, 7) \), \( (8, 9, 11) \), and so on.
   - Let \( S_i \) denote the set containing the \( i \)-th term of each group. Thus, \( S_1 \) contains the first term of each group, \( S_2 \) contains the second term, and so on up to \( S_k \).

2. **Define the function \( f \):**
   - For \( n \) indivisible by 5, there exists a unique \( t \) and \( m \) such that \( n = S_t(m) \).
   - Define \( f(n) \) as follows:
     \[
     f(n) = 
     \begin{cases} 
     S_{t+1}(m) & \text{if } t < k \\
     5S_1(m) & \text{if } t = k 
     \end{cases}
     \]
   - For \( n \) divisible by 5, let \( v_5(n) \) be the highest power of 5 dividing \( n \). Define:
     \[
     f(n) = 5^{v_5(n)} f\left(\frac{n}{5^{v_5(n)}}\right)
     \]
   - Define \( f(0) = 0 \).

3. **Verify the function properties:**
   - For \( n \) indivisible by 5, \( f^k(n) = 5n \) by construction.
   - For \( n \) divisible by 5, the function \( f \) reduces the problem to a smaller \( n \) until it becomes indivisible by 5, at which point the previous case applies.

4. **Check the specific case \( k = 2003 \):**
   - Setting \( k = 2003 \), we need to verify that \( f^{2003}(n) = 5n \) for all \( n \in \mathbb{N}_0 \).
   - By the construction, \( f^{2003}(n) = 5n \) holds for all \( n \) indivisible by 5.
   - For \( n \) divisible by 5, the function \( f \) ensures that the property holds recursively.

Thus, the function \( f \) defined above satisfies \( f^{2003}(n) = 5n \) for all \( n \in \mathbb{N}_0 \).

The final answer is \( \boxed{ f^{2003}(n) = 5n } \) for all \( n \in \mathbb{N}_0 \).