Let $\phi(n)$ denote the number of positive integers less than or equal to $n$ and relatively prime to $n$. Find all natural numbers $n$ and primes $p$ such that $\phi(n)=\phi(np)$.
1. Let \( n \) be a natural number and \( p \) be a prime such that \(\phi(n) = \phi(np)\). Recall that \(\phi(n)\) is the Euler's totient function, which counts the number of positive integers up to \( n \) that are relatively prime to \( n \).

2. If \( n = \prod_{i=1}^k p_i^{q_i} \) for distinct primes \( p_i \), then the Euler's totient function is given by:
   \[
   \phi(n) = n \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right)
   \]
   This can be rewritten as:
   \[
   \phi(n) = \prod_{i=1}^k (p_i - 1) p_i^{q_i - 1}
   \]

3. Now, consider \( np \). Since \( p \) is a prime and \( n \) is composed of primes \( p_i \), we have:
   \[
   np = \prod_{i=1}^k p_i^{q_i} \cdot p
   \]
   The Euler's totient function for \( np \) is:
   \[
   \phi(np) = \phi(n) \cdot \phi(p) = \phi(n) \cdot (p - 1)
   \]

4. Given that \(\phi(n) = \phi(np)\), we have:
   \[
   \phi(n) = \phi(n) \cdot (p - 1)
   \]
   Dividing both sides by \(\phi(n)\) (assuming \(\phi(n) \neq 0\)), we get:
   \[
   1 = p - 1
   \]
   Therefore:
   \[
   p = 2
   \]

5. Now, we must ensure that \( n \) is not divisible by 2. If \( n \) were divisible by 2, then \(\phi(n)\) would be multiplied by \( \frac{1}{2} \) when considering \(\phi(np)\), which contradicts our initial condition. Therefore, \( n \) must be odd.

6. Hence, the only solution is when \( p = 2 \) and \( n \) is any odd natural number.

The final answer is \( \boxed{ p = 2 } \) and \( n \) is any odd natural number.