1. Let $\mathbf{P}$ be the set of all prime numbers. Find all functions $f: \mathbf{P} \rightarrow \mathbf{P}$ such that for any $p, q \in \mathbf{P}$, we have
$$
(f(p))^{f(q)} + q^{p} = (f(q))^{f(p)} + p^{q} .
$$
1. Clearly, for any $p \in \mathbf{P}, f(p)=p$ satisfies the requirement.
Now we prove that this is the only solution.
First, we prove that $f(2)=2$.
Substitute $q=2$ into the condition equation to get
$$
(f(p))^{f(2)}+2^{p}=(f(2))^{f(p)}+p^{2} .
$$

If $f(2) \neq 2$, then for any odd prime $p$, the right side of the above equation is even. Therefore, it must be that $f(p)=2$. In this case, for any two different odd primes $p$ and $q$, we have
$$
2^{2}+q^{p}=p^{q}+2^{2} \Rightarrow p=q,
$$

which is a contradiction.
Thus, $f(2)=2$.
Therefore, for any odd prime $p$, we have
$$
\begin{array}{l}
(f(p))^{2}+2^{p}=2^{f(p)}+p^{2} \\
\Rightarrow 2^{p}-p^{2}=2^{f(p)}-(f(p))^{2} .
\end{array}
$$

Let $g(n)=2^{n}-n^{2}$, then for any positive integer $n \geqslant 3$, we have
$$
g(n+1)-g(n)=2^{n}-2 n-1>0 .
$$

Hence, $g(n)$ is strictly increasing on the set $\{3,4, \cdots\}$. Therefore, equation (1) indicates that $f(p)=p$ for any odd prime, and together with $f(2)=2$, we get that for any $p \in \mathbf{P}$, $f(p)=p$, which is the unique solution to the problem.