Is the identity function a unique multiplicative homeomorphism of $\mathbb N$? 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 $x,y\in\mathbb N$.

Problem 1. Is the identity function a unique multiplicative homeomorphism of $\mathbb N_\tau$?

An affirmative answer to this problem follows from an affirmative answer to

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 $\;\;b^n\not\equiv 1\!\!\!\mod\! c$?

 A: 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.
