11. (Mathematics for Middle School, 1994, Issue 4, Olympiad Training Problem) Find the real-coefficient polynomial $f(x)$ that satisfies the following conditions: (1) For any real number $a$, $f(a+1)=f(a)+f(1)$; (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 $f(x)$.
11. $f(a+1)=f(a)+f(1)$, let $a=0 \Rightarrow f(1)=f(0)+f(1) \Rightarrow f(0)=0$, so $f(x)=a_{1} x+$ $a_{2} x^{2}+\cdots+a_{n} x^{n}$, and by the condition $f(x+1)=f(x)+f(1)$, we get
$$
h(x)=a_{1}(1+x)+a_{2}(1+x)^{2}+\cdots+a_{n}(1+x)^{n}=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+a_{1}+a_{2}+\cdots+a_{n}=g(n) .
$$

Taking the $n$-th derivative on both sides,
$$
h^{(n)}(x)=a_{n} \cdot n \cdot(n-1) \cdots 2 \cdot 1=g^{(n)}(x) \Rightarrow a_{n}=0 \text {, }
$$

Thus, the equation becomes
$$
h(x)=a_{1}(1+x)+a_{2}(1+x)^{2}+\cdots+a_{n-1}(1+x)^{n-1}=a_{1} x+a_{2} x^{2}+\cdots+a_{n-1} x^{n-1}+a_{1}+a_{2}+\cdots+
$$
$a_{n-1}=g(x)$. Taking the $(n-1)$-th derivative on both sides, similarly we get $a_{n-1}=0$, hence $a_{2}=a_{3}=a_{4}=\cdots=a_{n}=0$. The equation then becomes $a_{1}(1+x)=a_{1} x+a_{1}$, so $f(x)=a_{1} x$.
From the given condition (2), it is easy to get $f\left(k_{1}\right)=k_{1} \Rightarrow a_{1} k_{1}=k$.
Since $k_{1} \neq 0$, we have $a_{1}=1$. Therefore, the required polynomial is $f(x)=x$.