Example 7 Given $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, and satisfies the conditions:
(1) For any $x, y \in \mathbf{R}_{+}$, $f(x f(y))=y f(x)$;
(2) As $x \rightarrow+\infty$, $f(x) \rightarrow 0$.

Try to find the function $f(x)$.
Solution: Let $x=y$, then we have $f(x f(x))=x f(x)$.
Obviously, $x f(x)$ is a fixed point of $f(x)$, and the set of fixed points is $\left\{x f(x) \mid x \in \mathbf{R}_{+}\right\}$.
Now let $x=y=1$, substituting into condition (1) we get
$f(f(1))=f(1)$.
Next, let $x=1, y=f(1)$, substituting into the original equation we get $f(f(f(1)))=f^{2}(1)$.
Therefore, $f^{2}(1)=f(1)$.
Hence $f(1)=1$ (we discard $f(1)=0$), which means $x=1$ is a fixed point of $f(x)$. Thus, $x f(x)=1$.
So, $f(x)=\frac{1}{x}\left(x \in \mathbf{R}_{+}\right)$.
The following is a proof by contradiction that $x=1$ is the only fixed point of $f(x)$.
Assume there exists $\alpha \neq 1$ and $\alpha=f(\alpha)$.
(1) If $\alpha>1$, by $f(x f(x))=x f(x)$, let $x=\alpha$, we get $f(\alpha f(\alpha))=\alpha f(\alpha)$. Hence
$f\left(\alpha^{2}\right)=\alpha^{2}$.
Therefore, $f\left(\alpha^{4}\right)=\alpha^{4}, \cdots, f\left(\alpha^{2 n}\right)=\alpha^{2 n}$.
This contradicts $x \rightarrow+\infty, f(x) \rightarrow 0$.
(2) If $0<\alpha<1$,
Similarly, we derive a contradiction.
Thus, $f(x)$ has only one fixed point.