Endow the set $\mathbb N$ of positive integers with the topology $\tau$ generated by the base consisting of arithmetic progressions $a+b\mathbb N\_0$ where $\mathbb N\_0=\{0\}\cup\mathbb N$, where $a,b\in\mathbb N$. This topology is often referred to as the *Furstenberg topology* or the *profinite topology*. The space $\mathbb N\_\tau:=(\mathbb N,\tau)$ is homeomorphic to the space $\mathbb Q$ of rational numbers (being a second-countable regular countable space without isolated points). So $\mathbb N\_\tau$ has many non-trivial homeomorphisms. But I know no non-trivial homeomorphism of $\mathbb N\_\tau$ which would be *multiplicative* is the sense that $f(x\cdot y)=f(x)\cdot f(y)$ for any $x,y\in\mathbb N$.



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> **Problem 1.** Is the identity function a unique multiplicative homeomorphism of $\mathbb N\_\tau$?
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An affirmative answer to this problem follows from an affirmative answer to



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> **Problem 2.** Are there prime numbers $a,b,c$ such that for any $d\in\mathbb N$ there exists $n\in\mathbb N$ such that $a^n\equiv 1\!\!\!\mod\! d\;\;$ but $\;\;b^n\not\equiv 1\!\!\!\mod\! c$?
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Is the identity function a unique multiplicative homeomorphism of $\mathbb N$?
No.


First observe that the automorphisms of the semigroup $\mathbf{N}^\*$ (which you denote $\mathbb{N}$) are induced by permutations of primes.


Consider the automorphism $f$ induced by the transposition $(2,3)$ (thus, mapping $2^a.3^b.c$ to $2^b.3^a.c$, $c$ coprime to 6).


I claim that $f$ is continuous. Indeed, consider any convergent sequence $m\_i\to m$. Thus, $m\_i=m+r\_i$ with $r\_i$ tending to 0 in the profinite completion of $\mathbf{Z}$ (that is, for every $n\ge 1$ there exists $i\_0$ such that $n$ divides $r\_i$ for all large $i$).


Write $m=2^a.3^b.c$ with $c$ coprime to 6. There exists $i\_0$ such that $2^{a+1}3^{b+1}$ divides $r\_i$ for all $i\ge i\_0$. So, for $i\ge i\_0$, $m\_i=2^a.3^b.c+2^{a+1}3^{b+1}.t\_i$ for some $t\_i$. So $m\_i=2^a3^b(c+6t\_i)$ with $c$ coprime to 6.
Thus $f(m\_i)=2^b3^a(c+6t\_i)=f(m)+r'\_i$, with $r'\_i=2^{b-a}3^{a-b}r\_i$. 
Thus $r'\_i$ also tends to 0 in the profinite completion of $\mathbf{Z}$. Thus $f(m\_i)$ tends to $f(m)$. Hence, $f$ is continuous.


The argument adapts immediately to any finitely supported permutation of the set of primes.


