Example 7 Let $I=[0,1], G=\{(x, y) \mid x \in I, y \in I\}$, find all mappings $f$ from $G$ to $I$ such that for any $x, y, z \in I$ we have
(1) $f[f(x, y), z]=f[x, f(y, z)]$;
(2) $f(x, 1)=x, f(1, y)=y$;
(3) $f(z x, z y)=z^{k} \cdot f(x, y)$, where $k$ is a positive number independent of $x, y, z$.
(CMO - 6 Problem)
From conditions (3) and (2), we have
$$
\begin{array}{l}
f(x, y)=y^{k} \cdot f\left(\frac{x}{y}, 1\right)=y^{k-1} \cdot x \quad (0 \leqslant x \leqslant y, 0<y), \\
f(x, y)=x^{k} \cdot f\left(1, \frac{y}{x}\right)=x^{k-1} \cdot y \quad (0 \leqslant y \leqslant x, 0<x).
\end{array}
$$

Taking $0<x<y$, and $x$ sufficiently small, such that $y^{k-1} \cdot x<z, x<z^{k-1} \cdot y$. By conditions (1) and (2), we also have
$$
\begin{aligned}
z^{k-1} y^{k-1} x & =f\left(y^{k-1} \cdot x, z\right)=f[f(x, y), z] \\
& =f[x, f(y, z)]=f\left(x, z^{k-1} \cdot y\right) \\
& =\left(z^{k-1} \cdot y\right)^{k-1} \cdot x .
\end{aligned}
$$

This leads to $(k-1)(k-2)=0$, solving which gives $k_{1}=1, k_{2}=2$.
Substituting $k_{1}$ and $k_{2}$ into (1) and (2), we get
$$
\begin{aligned}
f_{1}(x, y) & =\left\{\begin{array}{l}
x, \text { when } 0 \leqslant x \leqslant y, 0<y, \\
y, \text { when } 0 \leqslant y \leqslant x, 0<x
\end{array}\right. \text { . } \\
f_{2}(x, y) & =x y, \text { when }(x, y) \neq(0,0).
\end{aligned}
$$

When $x=y=0$, by condition (3), we directly obtain
$$
f(0,0)=f(z \cdot 0, z \cdot 0)=z^{k} \cdot f(0,0),
$$

where $z \in I$ is arbitrary, hence it must be that $f(0,0)=0$.
This means that for the functions $f_{1}(x, y)$ and $f_{2}(x, y)$ given by (3) and (4), we should define $f_{1}(0,0)=0, f_{2}(0,0)=0$.
It is easy to verify: both functions satisfy conditions (1) to (3).