Find all $k>0$ such that there exists a function $f : [0,1]\times[0,1] \to [0,1]$ satisfying the following conditions:
$f(f(x,y),z)=f(x,f(y,z))$;
$f(x,y) = f(y,x)$;
$f(x,1)=x$;
$f(zx,zy) = z^{k}f(x,y)$, for any $x,y,z \in [0,1]$
1. **Given Conditions:**
   - \( f(f(x,y),z) = f(x,f(y,z)) \)
   - \( f(x,y) = f(y,x) \)
   - \( f(x,1) = x \)
   - \( f(zx,zy) = z^k f(x,y) \) for any \( x, y, z \in [0,1] \)

2. **Analyzing the conditions:**
   - From \( f(x,1) = x \), we know that \( f(1,1) = 1 \).
   - Let \( P(x,y) \) denote the assertion \( f(f(x,y),z) = f(x,f(y,z)) \).

3. **Finding \( f(x,x) \):**
   - Consider \( P(x,x) \):
     \[
     f(x,x) = f(x \cdot 1, x \cdot 1) = x^k f(1,1)
     \]
     Since \( f(1,1) = 1 \), we have:
     \[
     f(x,x) = x^k \quad \text{(1)}
     \]

4. **Using \( f(f(x,x),f(x,x)) \):**
   - From (1), we have \( f(x,x) = x^k \).
   - Consider \( P(f(x,x),f(x,x)) \):
     \[
     f(f(x,x),f(x,x)) = f(x^k, x^k) = (x^k)^k = x^{k^2}
     \]
   - Also, using the functional equation:
     \[
     f(f(x,x),f(x,x)) = f(f(f(x,x),x),x)
     \]
     Using \( f(x,x) = x^k \):
     \[
     f(f(x^k,x),x) = f(x^k \cdot f(x^{k-1},1), x) = f(x^k \cdot x^{k-1}, x) = f(x^{2k-1}, x)
     \]
     Using the scaling property:
     \[
     f(x^{2k-1}, x) = x^k f(x^{2k-2}, 1) = x^k \cdot x^{2k-2} = x^{3k-2}
     \]
   - Equating both expressions for \( f(f(x,x),f(x,x)) \):
     \[
     x^{k^2} = x^{3k-2}
     \]
     Since this must hold for all \( x \in [0,1] \), we equate the exponents:
     \[
     k^2 = 3k - 2
     \]
     Solving the quadratic equation:
     \[
     k^2 - 3k + 2 = 0
     \]
     \[
     (k-1)(k-2) = 0
     \]
     \[
     k = 1 \quad \text{or} \quad k = 2
     \]

5. **Verifying the solutions:**
   - For \( k = 1 \):
     \[
     f(x,y) = \min\{x,y\}
     \]
     This satisfies all given conditions.
   - For \( k = 2 \):
     \[
     f(x,y) = xy
     \]
     This also satisfies all given conditions.

Conclusion:
The values of \( k \) that satisfy the given conditions are \( k = 1 \) and \( k = 2 \).

The final answer is \( \boxed{k = 1, 2} \).