Let $\mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)}$ be defined as
$$
a_{ij} = c_{i+j \operatorname{mod} (n+1)},
$$
so
$$
\mathbf A = \begin{pmatrix}
c_0 & c_1 & \dots & c_n\\
c_1 & c_2 & \dots & c_0\\
\vdots & \vdots & \ddots & \vdots\\
c_n & c_0 & \dots & c_{n-1}
\end{pmatrix}.
$$
Let's assume every index is taken modulo $n+1$.
Then for $\mathbf y = \mathbf A \mathbf x$
$$
y_m = \sum_{j=0}^n a_{mj} x_j = \sum_{j = 0}^n c_{m+j} x_j
$$
Let's use the following discrete Fourier transform
$$
\mathcal{F}[\mathbf y]_k = \sum_{m=0}^n y_m \exp\left\{-2\pi i\frac{k m}{n+1}\right\}
$$
Then
$$
\mathcal{F}[\mathbf y]_k = \sum_{m=0}^n \sum_{j=0}^n c_{m+j}x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\
= \sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=0}^n c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\
=\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=-j}^{n-j} c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} =\\
=\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m'=0}^{n} c_{m'} \exp\left\{2\pi i\frac{k m')}{n+1}\right\} = \\
=\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\mathcal{F}[\mathbf c]_{k} = \mathcal{F}^{*}[\mathbf x^*]_k \mathcal{F}[\mathbf c]_{k}
$$
Thus for $\mathbf A\mathbf x = \lambda \mathbf x$
$$
\mathcal{F}[\mathbf c]_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k
$$
Let for simplicity $C_k = \mathcal{F}[\mathbf c]_{k}$.
$$
C_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k
$$
Recall that discrete Fourier transform can be written in form
$$
\mathcal{F}[\mathbf x]_k = (\mathbf F\mathbf x)_k
$$
with $\mathbf F^{-1} = \frac{1}{n+1}\mathbf F^H$. Using $\mathbf C = \operatorname{diag} C_k$ one can write the eigenproblem as
$$
\mathbf C \mathbf F^* \mathbf x = \lambda \mathbf F \mathbf x
$$
The eigenvalues satisfy
$$
\operatorname{det}(\mathbf C \mathbf F^* - \lambda \mathbf F) = 0\\
\operatorname{det}(\mathbf C - \lambda\frac{1}{n+1} \mathbf F\mathbf F^\top) = 0
$$
The $\frac{1}{n+1} \mathbf F\mathbf F^\top$ has the following form
$$
\frac{1}{n+1} \mathbf F\mathbf F^\top =
\begin{pmatrix}
1 & 0 & \dots & 0 & 0\\
0 & 0 & \dots & 0 & 1\\
0 & 0 & \dots & 1 & 0\\
\vdots & \vdots & \cdot & \vdots & \vdots\\
0 & 1 & \dots & 0 & 0
\end{pmatrix}
$$
and the eigenvalues (except for $\lambda = C_0$) are the solutions to the problem
$$
\left|
\mathbf C_\text{AC} - \lambda \mathbf J_n
\right| = 0\\
\left|
\mathbf J_n \mathbf C_\text{AC} - \lambda \mathbf I_n
\right| = 0
$$
where $\mathbf J_n$ is an exchange matrix of size $n$. We arrived to a problem of finging the eigenvalues of the antidiagonal matrix $\mathbf J_n \mathbf C_\text{AC}$.
Next, following this post one can note that
$$
(\mathbf J_n \mathbf C_\text{AC})^2 = \operatorname{diag}
\begin{pmatrix}C_1 C_n & C_2 C_{n-1} & \dots & C_n C_1\end{pmatrix}
$$
so eigenvalues of the original problem are
$$
\lambda_0 = C_0\\
\lambda_{i, n-i+1} = \pm \sqrt{C_i C_{n-i+1}},\quad i = 1, 2, \dots
$$
