is there a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that 
$i) \exists n\in \mathbb{N}:f(n)\neq n$ 
$ii)$ the number of divisors of $m$ is $f(n)$  if and only if the number of divisors of $f(m)$ is $n$
1. **Define the function \( d(m) \) and the sets \( T(n) \):**
   - Let \( d(m) \) be the number of divisors of \( m \).
   - Let \( T(n) \) be the set of numbers \( x \) such that \( d(x) = n \).

2. **Establish the relationship:**
   - We need to find a function \( f \) such that \( d(m) = f(n) \) if and only if \( d(f(m)) = n \).

3. **Initial values:**
   - Define \( f(1) = 1 \) and \( f(2) = 2 \).

4. **Define the iteration function \( h(n) \):**
   - For \( n \geq 2 \), define \( h(n) \) as the number of iterations needed to reach 2 by repeatedly applying \( d \). Let \( h(2) = 0 \).

5. **Define the sets \( X_{ki_1...i_k} \):**
   - For all \( k \geq 0 \) and positive integers \( i_1, \ldots, i_k \), define the set \( X_{ki_1...i_k} \) recursively:
     - \( X_{0} \) is the set of numbers with \( d(x) = 2 \) (i.e., prime numbers).
     - \( X_{ki_1...i_k} \) is the set of numbers that are the \( i_k \)-th smallest element of \( T(x) \), where \( x \) runs across \( X_{(k-1)i_1...i_{k-1}} \).

6. **Assign \( f \) values within \( X \)-sets:**
   - Each \( n \geq 3 \) belongs to exactly one \( X \)-set.
   - For \( n \in X_0 \) (prime numbers), define \( f \) as a bijection. For example, order the odd prime numbers as \( \ldots, p_{-1}, p_0, p_1, \ldots \) and let \( f(p_i) = p_{i+1} \).

7. **Define \( f \) for other \( X \)-sets:**
   - For \( n \in X_{ki_1...i_k} \), consider \( f \) in \( X_{(k-1)i_1...i_{k-1}} \).
   - If \( f(m) = d(n) \), then there is exactly one number \( m' \) in \( X_{ki_1...i_k} \) such that \( d(m') = m \). Define \( f(n) = m' \).

8. **Verify the properties:**
   - \( f \) is a bijection and well-defined.
   - \( f(3) \neq 3 \).
   - If \( d(m) = f(n) \), then by definition, \( f(m) = n' \) where \( d(n') = n \).
   - If \( d(f(m)) = n \), then by definition \( d(m) = f(n) \), by the recursive definition of the \( X \)-sets.

Thus, such a function \( f \) does exist.

The final answer is \( \boxed{ f } \) exists.