1.6.5 ** Find all functions \( f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*} \), such that for any \( x, y \in \mathbf{N}^{*} \), we have
$$
f(x+f(y))=f(x)+y .
$$
Let $f(1)=a$, then from equation (1) we know $f(1+f(1))=f(1)+1$, i.e., $f(1+a)=1+a$, and thus $f(1+f(1+a))=f(1)+1+a$, i.e., $f(2+a)=2a+1$. On the other hand, $f(2+a)=f(2+f(1))=f(2)+1$. Therefore, $f(2)=2a$. Consequently, we have
$$
f(1+2a)=f(1+f(2))=f(1)+2=a+2,
$$

and also
$$
f(1+2a)=f(a+f(1+a))=f(a)+1+a,
$$

comparing the above two equations, we get $f(a)=1$. Thus, we have $f(2a)-f(a+f(1))=f(a)+1=2$, so
$$
2=f(1+2a-1)-f(2a-1+f(a))=a+f(2a-1).
$$

Given $f(2a-1) \in \mathbf{N}^{*}$, it can only be $a-1$, i.e., $f(1)-1$. Therefore, from equation (1), we know that for any $x \in \mathbf{N}^{*}$, we have $f(x+1)=f(x)+1$.

Using $f(1)-1$ and mathematical induction, it is easy to prove: for any $x \in \mathbf{N}^{*}$, we have $f(x)-x$. And $f(x)=x$ clearly satisfies (1), thus, the function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ we are looking for is $f(x)=x$.