*Two, find the real-coefficient polynomial $f(x)$ that satisfies the following conditions:
(1) For any real number $a$, we have
$$
f(a+1)=f(a)+f(1) \text {; }
$$
(2) There exists a real number $k_{1} \neq 0$, such that $f\left(k_{1}\right)=k_{2}$, $f\left(k_{2}\right)=k_{3}, \cdots, f\left(k_{n-1}\right)=k_{n}, f\left(k_{n}\right)=k_{1}$, where $n$ is the degree of $f(x)$.
II. Taking $a=0$, from (1) we get $f(1)=f(0)+f(1)$. Thus, $f(0)=0$, and $f(x)$ does not contain a constant term. We denote
$$
f(x)=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}
$$

From condition (1), we know
$$
\begin{array}{l}
a_{1}(1+x)+a_{2}(1+x)^{2}+\cdots+a_{n}(1+x)^{n} \\
=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+a_{1}+a_{2}+\cdots+a_{n}
\end{array}
$$

By expanding and comparing the coefficients on both sides, we get
$$
a_{2}=a_{3}=\cdots=a_{n}=0 \text {, hence } f(x)=a_{1} x \text {. }
$$

From (2), we get $f\left(k_{1}\right)=a_{1} k_{1}=k_{1}$, but $k_{1} \neq 0$, so $a_{1}=1$. The required polynomial is $f(x)=x$.