**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 $G$?




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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](https://en.wikipedia.org/wiki/Greedy_coloring#Optimal_ordering)).



Graph with only one coloring bijection
For every graph, every optimal coloring of the graph, and every ordering of the colors of the graph, there is another optimal coloring (possibly the same) in which each color class is maximal among the remaining vertices not included in earlier color classes. For this maximal coloring, any ordering of the vertices consistent with the ordering of colors causes the greedy algorithm to reconstruct the coloring.


So, no, there cannot be only one coloring bijection in your sense. Even when there is only one optimal coloring, you can permute the color classes of the coloring arbitrarily, and permute the vertices within each color class.


