Given $I=[0,1]$ and $G=\{(x,y)|x,y \in I\}$, find all functions $f:G\rightarrow I$, such that $\forall x,y,z \in I$ we have:
i. $f(f(x,y),z)=f(x,f(y,z))$;
ii. $f(x,1)=x, f(1,y)=y$;
iii. $f(zx,zy)=z^kf(x,y)$.
($k$ is a positive real number irrelevant to $x,y,z$.)
To find all functions \( f: G \rightarrow I \) that satisfy the given conditions, we will analyze each condition step by step.

1. **Condition (i):** \( f(f(x,y),z) = f(x,f(y,z)) \)

   This condition suggests that \( f \) is associative. Associativity is a key property that will help us narrow down the possible forms of \( f \).

2. **Condition (ii):** \( f(x,1) = x \) and \( f(1,y) = y \)

   These conditions imply that \( 1 \) acts as an identity element for the function \( f \). This means that combining any element with \( 1 \) using \( f \) will return the original element.

3. **Condition (iii):** \( f(zx,zy) = z^k f(x,y) \)

   This condition involves a scaling property with a positive real number \( k \). We need to consider how this scaling affects the function \( f \).

Let's consider possible forms of \( f \) that satisfy these conditions.

### Case 1: \( k = 1 \)

If \( k = 1 \), then condition (iii) becomes:
\[ f(zx, zy) = z f(x, y) \]

Let's test the function \( f(x, y) = \min(x, y) \):

- **Condition (i):** 
  \[ f(f(x, y), z) = f(\min(x, y), z) = \min(\min(x, y), z) = \min(x, y, z) \]
  \[ f(x, f(y, z)) = f(x, \min(y, z)) = \min(x, \min(y, z)) = \min(x, y, z) \]
  Both sides are equal, so condition (i) is satisfied.

- **Condition (ii):**
  \[ f(x, 1) = \min(x, 1) = x \]
  \[ f(1, y) = \min(1, y) = y \]
  Both conditions are satisfied.

- **Condition (iii):**
  \[ f(zx, zy) = \min(zx, zy) = z \min(x, y) = z f(x, y) \]
  This condition is satisfied.

Thus, \( f(x, y) = \min(x, y) \) is a solution when \( k = 1 \).

### Case 2: \( k = 2 \)

If \( k = 2 \), then condition (iii) becomes:
\[ f(zx, zy) = z^2 f(x, y) \]

Let's test the function \( f(x, y) = xy \):

- **Condition (i):**
  \[ f(f(x, y), z) = f(xy, z) = (xy)z = xyz \]
  \[ f(x, f(y, z)) = f(x, yz) = x(yz) = xyz \]
  Both sides are equal, so condition (i) is satisfied.

- **Condition (ii):**
  \[ f(x, 1) = x \cdot 1 = x \]
  \[ f(1, y) = 1 \cdot y = y \]
  Both conditions are satisfied.

- **Condition (iii):**
  \[ f(zx, zy) = (zx)(zy) = z^2 xy = z^2 f(x, y) \]
  This condition is satisfied.

Thus, \( f(x, y) = xy \) is a solution when \( k = 2 \).

### Conclusion

The only solutions to the given problem are:
- \( f(x, y) = \min(x, y) \) for \( k = 1 \)
- \( f(x, y) = xy \) for \( k = 2 \)

The final answer is \( \boxed{ f(x, y) = \min(x, y) } \) for \( k = 1 \) and \( f(x, y) = xy \) for \( k = 2 \).