Every fiber bundle with Cantor set fiber is the suspension of a homeomorphism of the Cantor set. I've heard that every fiber bundle (over $\mathbb S^1$?) with Cantor set fiber is the suspension of a homeomorphism of the Cantor set. 


*

*Can someone explain the intuition behind the fact? 

*Is there a canonical construction for such homeomorphism?

 A: There is a stronger fact proven by Dennis Sullivan: Any orientable one dimensional lamination which is transversally a Cantor
set (=solenoidal 1-dimensional manifold) is the suspension of 
a homeomorphism of tha Cantor set.
Sullivan, Dennis
Solenoidal manifolds. (English summary)
J. Singul. 9 (2014), 203–205 
Verjovsky, Alberto Commentaries on the paper Solenoidal manifolds by Dennis Sullivan [MR3249058]. J. Singul. 9 (2014), 245–251.
A: Let $F\to E \stackrel{p}{\to} S^1=\mathbb{R}/\mathbb{Z}$ be our bundle with fiber $F$ homeomorphic to a Cantor set.
Pick a basepoint $x_0\in S^1$ and let $F_0 = p^{-1}(x_0)$ be the fiber over $x_0$. Let $\pi \colon S^1\times\mathbb{R} \to S^1$ be rotation action on the circle (so $\pi(x,t)=\lfloor x+t\rfloor$). The action $\pi$ lifts to a unique action $\tilde{\pi}\colon E\times\mathbb{R}\to E$ on our total space by the local product structure (and the total disconnectedness of our fiber).
We have an restricted $\mathbb{Z}$-action $m\colon E\times\mathbb{Z}\to E$ on $E$ given by $m(x,n)=\tilde{\pi}(x,n)$ and this will give us the map we need.
Our homeomorphism $h\colon F_0\to F_0$ on the distinguished fiber $F_0$ is $h(x)= m(x,1)$. This is readily seen to be a homeomorphism on $F_0$ because the inverse is just given by $h^{-1}(x)=m(x,-1)$.
That is, intuitively $h$ takes a point in the fiber $F_0$ over the basepoint and 'follows the path' around the circle until it hits $F_0$ again. This is anologous to the Poincaré first return map (on a suitable neighbourhood of $F_0$ which is a local product $F_0\times (0,1)$) or the monodromy action of a covering space.
