Example 2 Whether there is a mapping $f: \mathbf{R}^{+} \rightarrow \mathbf{R}$, satisfying
$$
f^{(6-1)}(x)=(\sqrt{x}+1)^{2} ?
$$
The mapping $f(x)=\left(\sqrt{x}+\frac{1}{64}\right)^{2}$ satisfies the requirement. In fact, $\sqrt{f(x)}$
$$
\begin{array}{l}
=\sqrt{x}+\frac{1}{64}, \sqrt{f^{(2)}(x)}=\sqrt{f(x)}+\frac{1}{64}=\sqrt{x}+\frac{2}{64}, \cdots, \\
\sqrt{f^{(k)}(x)}=\sqrt{x}+\frac{k}{64}, \cdots, \sqrt{f^{(64)}(x)}=\sqrt{x}+1 .
\end{array}
$$

More generally, $f(x)=g\left(g^{-1}(x)+b\right)$ satisfies
$$
f^{(n)}(x)=g\left(g^{-1}(x)+n b\right),
$$

where $g$ is a one-to-one correspondence, and $b$ is any constant. In fact,
$$
\begin{array}{c}
g^{-1}(f(x))=g^{-1}(x)+b, \\
g^{-1}\left(f^{(2)}(x)\right)=g^{-1}(f(x))+b=g^{-1}(x)+2 b, \\
\cdots \cdots \\
g^{-1}\left(f^{(n)}(x)\right)=g^{-1}(x)+n b .
\end{array}
$$