Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(2m+2n)=f(m)f(n)$ for all natural numbers $m,n$.
1. **Define the assertion**: Let \( P(m, n) \) be the assertion \( f(2m + 2n) = f(m)f(n) \).

2. **Initial substitution**: Consider \( P(1, 1) \):
   \[
   P(1, 1) \implies f(2 + 2) = f(1)f(1) \implies f(4) = f(1)^2
   \]

3. **Introduce a constant \( c \)**: Let \( c = \frac{f(2)}{f(1)} \). This is valid as long as \( f(1) \neq 0 \).

4. **Substitute \( P(2, x) \)**:
   \[
   P(2, x) \implies f(2 \cdot 2 + 2x) = f(2)f(x) \implies f(4 + 2x) = f(2)f(x)
   \]
   Using \( f(2) = cf(1) \), we get:
   \[
   f(4 + 2x) = cf(1)f(x)
   \]

5. **Substitute \( P(1, x+1) \)**:
   \[
   P(1, x+1) \implies f(2 \cdot 1 + 2(x+1)) = f(1)f(x+1) \implies f(2 + 2x + 2) = f(1)f(x+1) \implies f(4 + 2x) = f(1)f(x+1)
   \]

6. **Equate the two expressions for \( f(4 + 2x) \)**:
   \[
   cf(1)f(x) = f(1)f(x+1)
   \]
   Since \( f(1) \neq 0 \), we can divide both sides by \( f(1) \):
   \[
   cf(x) = f(x+1)
   \]

7. **Solve the recurrence relation**: The relation \( f(x+1) = cf(x) \) suggests that \( f \) is a geometric sequence. Assume \( f(1) = a \), then:
   \[
   f(x) = a c^{x-1}
   \]

8. **Substitute back into the original equation**: Substitute \( f(x) = a c^{x-1} \) into the original functional equation \( f(2m + 2n) = f(m)f(n) \):
   \[
   f(2m + 2n) = a c^{2m + 2n - 1}
   \]
   \[
   f(m)f(n) = (a c^{m-1})(a c^{n-1}) = a^2 c^{m+n-2}
   \]
   Equate the two expressions:
   \[
   a c^{2m + 2n - 1} = a^2 c^{m+n-2}
   \]
   Divide both sides by \( a c^{m+n-2} \):
   \[
   c^{m+n-1} = a c^{m+n-2}
   \]
   Since \( c \neq 0 \), divide both sides by \( c^{m+n-2} \):
   \[
   c = a
   \]

9. **Determine the constants**: From \( c = a \), we have \( f(1) = a \) and \( f(2) = a^2 \). Since \( c = a \), we get \( a = 1 \). Therefore, \( f(1) = 1 \) and \( f(2) = 1 \).

10. **Conclude the function**: Since \( f(x) = a c^{x-1} \) and \( a = 1 \), \( c = 1 \), we get:
    \[
    f(x) = 1 \cdot 1^{x-1} = 1
    \]

Thus, the unique solution is:
\[
\boxed{f(x) = 1 \text{ for all } x \in \mathbb{N}}
\]