Consider an $n\times n$ matrix $A$ defined over $\mathbb{F}$ where $1\neq 0$. This matrix is given by: \begin{equation} \begin{pmatrix} a_{1} & a_{2} & a_{3} & \ldots & a_{n-1} & a_{n} \\ a_{n} & a_{1} & a_{2} &\ldots & a_{n-2} & a_{n-1}\\ \vdots & \vdots & \ddots & & \vdots & \vdots\\ \vdots & \vdots & & \ddots & \vdots & \vdots\\ a_{3} & a_{4} & a_{5} & \ldots & a_{1} & a_{2} \\ a_{2} & a_{3} & a_{4} & \ldots & a_{n} & a_{1} \end{pmatrix} \end{equation} For $n=1$, $A$ is invertible if and only if $a_1\neq 0$.
For $n=2$, $A$ is invertible $\Longleftrightarrow\det A=a_1^2-a_2^2\neq0\Longleftrightarrow a_1\neq \pm a_2$.
When $a_2=a_3=\ldots =a_n$, we can calculate $\det A=\left[a_1+(n-1)a_2\right](a_1-a_2)^{n-1}$.Therefore $A$ is invertible if and only if $a_1\neq -(n-1)a_2$ and $a_1\neq a_2$.
If $\mathbb{F}$ is a field with a norm (for example, $\mathbb{R}$ or $\mathbb{C}$) denoted as $|\cdot|$, we can establish that $A$ is invertible if there exists an integer $i\in \{1,2,\ldots,n\}$ such that $2|a_i|>\sum^{n}_{j=1}|a_j|$.
However, finding a general condition seems to be quite challenging.