The Golomb space $\mathbb G$ is the set of positive integers endowed with the topology generated by the base consisting of the arithmetic progressions $a+b\mathbb N_0$ with relatively prime $a,b$ and $\mathbb N_0=\{0\}\cup\mathbb N$. It is known that the space $\mathbb G$ is connected and Hausdorff.
It is also easy to check that the multiplication map $\cdot:\mathbb G\times \mathbb G\to\mathbb G$, $\cdot:(x,y)\mapsto xy$, is continuous, so $\mathbb G$ is a commutative topological semigroup.
Problem. Is the Golomb space $\mathbb G$ topologically homogeneous? Or maybe rigid?
We recall that a topological space $X$ is rigid if its homeomorphism group is trivial.
This problem was motivated by this question, which discusses the relation of the Golomb space to another countable connected Hausdorff space, called the rational projective space $\mathbb QP^\infty$. This space is easily seen to be topologically homogeneous.
Best Answer
[Edit, Dec 6, 2019] I have a pleasure to inform that this problem was finally resolved in affirmative by T.Banakh, D.Spirito and S.Turek who proved the following
Edit: now published: The Golomb space is topologically rigid. Comment. Math. Univ. Carolin. 62 (2021), no. 3, 347–360.