6・163 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) $f(z x, z y)=z^{k} f(x, y)$, where $k$ is a positive number independent of $x, y, z$.
[Solution] From conditions (3) and (2), we have
$$\begin{array}{l}
f(x, y)=y^{k} f\left(\frac{x}{y}, 1\right)=y^{k-1} x, 0 \leqslant x \leqslant y, 00 ; \\
y, \text { when } 0 \leqslant y \leqslant x, x>0 .
\end{array}\right. \\
f_{2}(x, y)=x y, \text { when }(x, y) \neq(0,0) .
\end{array}$$

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

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 supplement the definitions $f_{1}(0,0)=0, f_{2}(0,0)=0$.

It is easy to verify that both functions satisfy conditions (1) and (3).