Nonrigidity for circle homeomorphisms with several break points

Let $f$ and $g$ be two class $P$-homeomorphisms of the circle $S^1$ with break points singularities. Assume that the derivatives $\text{Df}$ and $\text{Dg}$ are absolutely continuous on every continuity interval of $\text{Df}$ and $\text{Dg}$ respectively. Denote by $C(f)$ the set of break points of $f$. For $c\in S^1$, denote by ${\pi }_{s,O_f(c)}(f)$ the product of $f-$ jumps in break points lying to the $f-$ orbit of $c$ and by $$\text{SO}$(f) = {O_{f}©:~c \in C(f)~$\text{and}$~\pi_{s, O_{f}©}(f)\neq 1}$,calledthesetofsingular$f$-orbits\mathrm{.}Themaps$f$and$g$ are called break-equivalent if there exists a topological conjugating $h$ such that $h($\text{SO}$(f))=$\text{SO}$(g)~~ $\text{and}$ ~~ \pi_{s, O_{g}(h©)}(g) = \pi_{s, O_{f}©}(f) ~~ $\text{for all}$~~ c\in $\text{SO}$(f)$\mathrm{.}Assumethat$f$and$g$ have the same irrational rotation number of bounded type. We prove that if $f$ and $g$ are not break-equivalent, then any topological conjugating $h$ between $f$ and $g$ is a singular function i.e. it is a continuous on $S^1$, but $\text{Dh}(x)=0$ a.e. with respect to the Lebesgue measure. As a consequence if for some point $d\in \text{SO}(f)$, $\pi_{s, O_{g}(d)}(g)\notin {\pi_{s, O_{f}©}(f): c\in C(f)}$, then the homeomorphism conjugation $h$ is a singular function. This later result generalizes previous results for one and two break points obtained by Dzhalilov-Akin-Temir and Akhadkulov-Dzhalilov-Noorani. Moreover, if $f$ and $g$ do not have the same number of singular orbits then the homeomorphism conjugating $f$ to $g$ is a singular function.