Prove that two $n$-sheeted covering space of $S^{1}$ are isomorphic.

Question

We have a Blaschke product $B(z) \colon S^1 \to S^1$ of order $n$ and the map $f \colon S^1 \to S^1$, $f(z)=z^n$. Both maps are regular on $S^1$.

We have already proved that both are $n$-sheeted covering spaces. How could we prove that they are isomorphic, this means that there exists a homeomorphism $c$, such that $B \circ c = f$ (or $f \circ c = B$) on $S^1$

a) without the fundamental group (more important!)

b) with the fundamental group (probably using the theorem if the groups are conjugate, the covering spaces are isomorphic + the fact that both spaces have the same fundamental group?)

Answer 1

Can we just take $c=B^{\frac{1}{n}}$ (then $f \circ B^{\frac{1}{n}} = B$) ?

$B$ is nonzero and holomorphic, so the $n$-th root exists (is there a problem since $S^1$ is not simply connected?)

Then $B^{\frac{1}{n}}$ is clearly continous and $S^1$ is clearly a compact and Hausdorff space. But is the $n$-th root here also bijective ? If it is and the above is true, then it should be a homeomorphism.

Is there a flaw in my reasoning/are any of the assumptions false/why are they correct ?

Thanks