Let $\mathbb{N}^*=\{1,2,\ldots\}$. Find al the functions $f: \mathbb{N}^*\rightarrow \mathbb{N}^*$ such that:

(1) If $x<y$ then $f(x)<f(y)$.
(2) $f\left(yf(x)\right)=x^2f(xy)$ for all $x,y \in\mathbb{N}^*$.
1. **Given Conditions:**
   - \( f: \mathbb{N}^* \rightarrow \mathbb{N}^* \)
   - If \( x < y \), then \( f(x) < f(y) \).
   - \( f(yf(x)) = x^2 f(xy) \) for all \( x, y \in \mathbb{N}^* \).

2. **Assume \( f(y) = y^2 \) for \( y \) of the form \( xf(x) \):**
   - We need to show that \( f(y) = y^2 \) for any \( y \).

3. **Assume \( f(y) > y^2 \):**
   - By the given condition (1), \( f(f(y)) > f(y^2) = f(y)^2 \).
   - Using the functional equation (2), \( f(y)^2 < f(f(y)) = y^2 f(y) \).
   - This implies \( f(y) < y^2 \), which is a contradiction.

4. **Assume \( f(y) < y^2 \):**
   - Similarly, by the given condition (1), \( f(f(y)) < f(y^2) = f(y)^2 \).
   - Using the functional equation (2), \( f(y)^2 > f(f(y)) = y^2 f(y) \).
   - This implies \( f(y) > y^2 \), which is also a contradiction.

5. **Conclusion:**
   - Since both assumptions \( f(y) > y^2 \) and \( f(y) < y^2 \) lead to contradictions, the only possibility is \( f(y) = y^2 \).

6. **Verification:**
   - Substitute \( f(y) = y^2 \) into the original functional equation:
     \[
     f(yf(x)) = f(yx^2) = (yx^2)^2 = y^2 x^4
     \]
     \[
     x^2 f(xy) = x^2 (xy)^2 = x^2 y^2 x^2 = y^2 x^4
     \]
   - Both sides are equal, confirming that \( f(y) = y^2 \) satisfies the functional equation.

The final answer is \( \boxed{ f(y) = y^2 } \).