11. Since $0,1,2, \cdots, n-1$ form a complete residue system modulo $n$, and $(k, n)=1$, it follows that $0, k, 2k, \cdots, (n-1)k$ is also a complete residue system modulo $n$. If we denote $a_{i} \equiv i k \pmod{n}, 0 \leqslant a_{i}<n$, then $\left\{a_{0}, a_{1}, \cdots, a_{n-1}\right\}=\{0,1,2, \cdots, n-1\}$. Therefore,
$$\left\{a_{1}, a_{2}, \cdots, a_{n-1}\right\}=\{1,2, \cdots, n-1\}=M$$

For any $i$, if $1 \leqslant i \leqslant n-2$, then
$$a_{i+1}-a_{i} \equiv (i+1)k - ik = k \pmod{n}$$

Since $0 < a_{i} < n$ and $0 < a_{i+1} < n$, it follows that $a_{i+1} = a_{i} + k$ or $a_{i+1} = a_{i} + k - n$. If $a_{i+1} = a_{i} + k$, then
$$a_{i} = a_{i+1} - k = \left|k - a_{i+1}\right|$$

By condition (2), $a_{i}$ and $a_{i+1}$ are the same color. If $a_{i+1} = a_{i} + k - n$, then
$$a_{i+1} = \left|a_{i} + k - n\right| = \left|k - (n - a_{i})\right|$$

By condition (2), $a_{i+1}$ and $n - a_{i}$ are the same color. If we further use (1), $n - a_{i}$ and $a_{i}$ are the same color, thus $a_{i+1}$ and $a_{i}$ are also the same color. By sequentially setting $i=1,2, \cdots, n-2$, we obtain the conclusion.