Unfortunately, my proof of the rigidity of the Golomb space contained a gap. So, only the first part of the problem can be answered (at the moment).

**Theorem.** The Golomb space $\mathbb G$ is not topologically homogeneous as $h(1)=1$ and $h(\Pi)=\Pi$ for any homeomorphism $h$ of $\mathbb G$.

Here $\Pi$ stands for the set of prime numbers.
The proof of this theorem will be divided into a sequence of 9 relatively simple and straightforward lemmas (whose detail proofs can be found here), but the last lemma exploits the powerful Dirichlet Theorem on primes in arithmetic progressions.

First let us fix notation. For a point $x\in\mathbb N$ by $\tau_x=\{U\in\tau:x\in U\}$ we denote the family of open neighborhoods of $x$ in the Golomb topology $\tau$ on $\mathbb N$.

For a number $x\in\mathbb N$ let $\Pi_x$ be the set of all prime divisors of $x$. Two numbers $x,y$ are *coprime* if $\Pi_x\cap\Pi_y=\emptyset$.

For a number $x\in\mathbb N$ and a prime number $p$ let $l_p(x)$ be the largest integer number such that $p^{l_p(x)}$ divides $x$. The function $l_p(x)$ plays the role of logarithm with base $p$.

A number $x$ is *square-free* if $l_p(x)\le 1$ for any prime number $p$.

**Lemma 1.** For a basic open set $a+b\mathbb N_0$ in $\mathbb G$ its closure
$$\overline{a+b\mathbb N_0}=\bigcap_{p\in\Pi_b}p\mathbb N\cup (a+p^{l_p(b)}\mathbb N).$$

Lemma 1 implies that the family $$\mathcal F_0=\{F\subset\mathbb N:\exists U_1,\dots,U_n\in\tau\setminus\{\emptyset\}\mbox{ such that }\bigcap_{i=1}^n\bar U_i\subset F\}$$is a filter on $\mathbb N$.

The definition of $\mathcal F_0$ implies that this filter is preserved by any homeomorphism $h$ of $\mathbb G$ (which means that the filter $h[\mathcal F_0]:=\{h(F):F\in\mathcal F_0\}$ coincides with $\mathcal F_0$).

**Lemma 2.** The filter $\mathcal F_0$ is generated by the base consisting of the sets $q\mathbb N$ with $q$ square-free.

**Lemma 3.** For every square-free number $q$ and any number $y\ne 1$ coprime with $q$ there are open sets $U_1\in\tau_1$ and $U_y\in\tau_y$ such that $\bar U_1\cap\bar U_y\subset q\mathbb N$.

**Lemma 4.** For every points $y\ne x\ne 1$, open sets $U_x\in\tau_x$, $U_y\in\tau_y$, and a prime number $p\in\Pi_x$, the set $\bar U_x\cap\bar U_y$ is not contained in $p\mathbb N$.

Lemmas 2-4 imply

**Lemma 5.** The number $1$ is a fixed point of any homeomorphism $h$ of the space $\mathbb G$.

For any point $x\in\mathbb N$ consider the filter $$\mathcal F_x=\{F\subset\mathbb N:\exists U_x\in\tau_x,\;\exists U_1\in\tau_1\mbox{ such that }\bar U_x\cap\bar U_1\subset F\}.$$

**Lemma 6.** For two numbers $x,y\in\mathbb N\setminus\{1\}$ the following conditions are equivalent:

$\bullet$ $\mathcal F_x\subset\mathcal F_y$;

$\bullet$ $\Pi_y\subset\Pi_x$.

Lemma 6 implies:

**Lemma 7.** For every homeomorphism $h$ of the Golomb space $\mathbb G$, and any numbers $x,y\in\mathbb N$ with $\Pi_x\subset\Pi_y$ we get $\Pi_{h(x)}\subset\Pi_{h(y)}$.

**Lemma 8.** For every homeomorphism $h:\mathbb G\to\mathbb G$ there exists a bijective map $\sigma:\Pi\to\Pi$ of the set $\Pi$ of all prime numbers such that $\Pi_{h(x)}=\sigma(\Pi_x)$ for any $x\in \mathbb N$.

Our final lemma completes the proof of the Theorem.

**Lemma 9.** $h(\Pi)=\Pi$ for any homeomorphism $h$ of $\mathbb G$.

*Proof.* It suffices to prove that $h(\Pi)\subset\Pi$. Let $\sigma:\Pi\to\Pi$ be the permutation from Lemma 8. Assuming that $h(\Pi)\not\subset\Pi$, we can find a prime number $p$ such that $h(p)\notin\Pi$. By Lemma 8, $h(p)=q^n$ for some $n>1$ and the prime number $q=\sigma(p)$. Taking into account that $\Pi_{h^{-1}(q)}=\{\sigma^{-1}(q)\}=\{p\}$ and $h^{-1}(q)\ne p=h^{-1}(q^n)$, we conclude that $h^{-1}(q)=p^k$ for some $k>1$. Now consider the continuous map $\gamma:\mathbb G\to\mathbb G$, $\gamma:x\mapsto x^n$, and the continuous map $f=h^{-1}\circ\gamma\circ h:\mathbb G\to\mathbb G$. Observe that $f(p^k)=h^{-1}\circ\gamma\circ h(p^k)=h^{-1}\circ\gamma(q)=h^{-1}(q^n)=p$. For the neighborhood $p+(p^k-1)\mathbb N_0$ of $p$, find a neighborhood $p^k+(p^k-1)b\mathbb N_0$ of $p^k$ such that $f(p^k+(p^k-1)b\mathbb N_0)\subset p+(p^k-1)\mathbb N_0$.

By the classical Dirichlet Theorem, the arithmetic progression $p^k+(p^k-1)b\mathbb N_0$ contains a prime number $r$. Lemma 8 implies that $f(r)=r^l$ for some $l\ge 1$. Taking into account that $r\in p^k+(p^k-1)b\mathbb N_0\subset 1+(p^k-1)\mathbb N_0$, we conclude that $f(r)=r^l=(p+(p^k-1)\mathbb N_0)\cap(1+(p^k-1)\mathbb N_0)^l\subset(p+(p^k-1)\mathbb N_0)\cap(1+(p^k-1)\mathbb N_0)=\emptyset$ and this is a desired contradiction.

**Remark.** An example of a rigid countable connected Hausdorff space was constructed by Joseph Martin.