First, we prove the uniqueness. If there exists another such mapping $g_{n}$, then when $m=e$, by equation (3) we get $g_{n}(e)=f_{n}(e)=n^{+}$. Assuming for some $m \in N$, $g_{n}(m)=f_{n}(m)$, then by equation (4) we get
$$g_{n}\left(m^{+}\right)=\left(g_{n}(m)\right)^{+}=\left(f_{n}(m)\right)^{+}=f_{n}\left(m^{+}\right)$$

Therefore, by Theorem 3 of §1 (the principle of mathematical induction), for all $m \in N$, we have $g_{n}(m)=f_{n}(m)$. This proves the uniqueness.

Next, we prove the existence. When $n=e$, we define the mapping from $N$ to itself as
$$f_{e}(m)=m^{+}, \quad m \in \mathbf{N}$$

It satisfies the requirements. In fact, since $f_{e}(e)=e^{+}$, condition (3) is satisfied. For any $m \in \boldsymbol{N}$, by definition (5) we get
$$f_{e}\left(m^{+}\right)=\left(m^{+}\right)^{+}=\left(f_{e}(m)\right)^{+},$$

which means condition (4) is satisfied. Assuming such a mapping $f_{n}$ exists for $n$, for $n^{+}$ we define the mapping from $\boldsymbol{N}$ to itself as
$$f_{n}^{+}(m)=\left(f_{n}(m)\right)^{+}, \quad m \in \mathbf{N}$$

From this and the fact that the mapping $f_{n}$ satisfies conditions (3) and (4), we derive
$$\begin{aligned} & f_{n^{+}}(e)=\left(f_{n}(e)\right)^{+}=\left(n^{+}\right)^{+} \\ f_{n^{+}}\left(m^{+}\right)= & \left(f_{n}\left(m^{+}\right)\right)^{+}=\left(\left(f_{n}(m)\right)^{+}\right)^{+}=\left(f_{n}^{+}(m)\right)^{+} \end{aligned}$$

This proves that the mapping $f_{n^{+}}$ satisfies conditions (3) and (4). Therefore, by the principle of mathematical induction (Theorem 3 of §1), for any $n \in \boldsymbol{N}$, there must exist a mapping $f_{n}$ that satisfies conditions (3) and (4). Proof completed.