In the system of base $n^2 + 1$ find a number $N$ with $n$ different digits such that:

[b](i)[/b] $N$ is a multiple of $n$. Let $N = nN'.$

[b](ii)[/b] The number $N$ and $N'$ have the same number $n$ of different digits in base $n^2 + 1$, none of them being zero.
[b]
(iii)[/b] If $s(C)$ denotes the number in base $n^2 + 1$ obtained by applying the permutation $s$ to the $n$ digits of the number $C$, then for each permutation $s, s(N) = ns(N').$
1. **Define the number \( N \) and \( N' \):**
   - We are given that \( N = nN' \).
   - \( N \) and \( N' \) should have \( n \) different digits in base \( n^2 + 1 \), none of them being zero.

2. **Express \( N \) and \( N' \) in base \( n^2 + 1 \):**
   - Let \( N = \sum_{i=1}^n i \cdot n \cdot (n^2 + 1)^{i-1} \).
   - In base \( n^2 + 1 \), this can be written as \( n, 2n, 3n, \ldots, nn \).
   - Let \( N' = \sum_{i=1}^n i \cdot (n^2 + 1)^{i-1} \).
   - In base \( n^2 + 1 \), this can be written as \( 1, 2, 3, \ldots, n \).

3. **Verify that \( N \) is a multiple of \( n \):**
   - By construction, \( N = nN' \), so \( N \) is clearly a multiple of \( n \).

4. **Verify that \( N \) and \( N' \) have \( n \) different digits:**
   - \( N \) has digits \( n, 2n, 3n, \ldots, nn \) in base \( n^2 + 1 \), which are \( n \) different digits.
   - \( N' \) has digits \( 1, 2, 3, \ldots, n \) in base \( n^2 + 1 \), which are \( n \) different digits.

5. **Verify the permutation property:**
   - For any permutation \( s \), \( s(N) = ns(N') \).
   - Since \( N = nN' \), applying the same permutation \( s \) to both \( N \) and \( N' \) will yield \( s(N) = n \cdot s(N') \).

Thus, the number \( N \) constructed as \( \sum_{i=1}^n i \cdot n \cdot (n^2 + 1)^{i-1} \) satisfies all the given conditions.

The final answer is \( \boxed{ N = \sum_{i=1}^n i \cdot n \cdot (n^2 + 1)^{i-1} } \).