**Question.** Is there a finite, simple undirected graph $G=(V,E)$ with more than $1$ vertex such that there is only $1$ coloring bijection (defined below) for $G$?

We denote by $\mathbb{N}$ the set of positive integers and set $[n] = \{1,\ldots,n\}$ for $n\in\mathbb{N}$.

Let $G= (V,E)$ be a simple undirected graph on $n\geq 1$ vertices, and let $b:[n]\to V$ be a bijection. We assign to $b$ the greedy coloring $c_b$ constructed by traversing the graph in the order $b$. Formally, with recursive definition of $c_b:[n] \to [n]$:

- $c_b(1) = 1$;
- if $k\in[n]$ and $k>1$ let $$c_b(k) = \min\:\big(\mathbb{N}\setminus\{c_b(j): j \in [k-1]\land \{b(j),b(k)\}\in E\}\big).$$

We call $b$ *coloring* if $\text{im}(c_b) = [\chi(G)]$. For every graph there is a coloring bijection (see here).