Consider the $(2q+1)\times(2q+1)$ ladder operator $$L=\left(\begin{matrix}0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots& \vdots&\cdots & \ddots&\vdots\\ 0&0&0&\cdots &1\\ 0&0&0&\cdots &0 \end{matrix}\right)$$ acting on $\mathbb{C}^{2q+1}$ . Given a vector $c\in\mathbb{C}^{2q+1}\quad$ with $\langle c,c \rangle<1$, find $d\in\mathbb{C}^{2q+1}\quad$ such that $\langle d,L^k\,d\rangle=-\langle c,L^k\,c\rangle$ for all $k=1,\cdots,2q$, where $\langle \cdot,\cdot\rangle$ is the standard inner product. Moreover, $\langle c,c \rangle+\langle d,d \rangle=1$.
I need to solve this problem for $d$ or at least find the conditions over $c$ such that it has a solution, but I don't even know where to start or to try finding something helpful.