Extending homeomorphisms in Cantor set Let $C'$ be the "endpoints" of the Cantor set $C$. These are the endpoints of the missing intervals, and we know $C'\simeq \mathbb Q$.
Does every homeomorphism $h:C'\to C'$ extend to a homeomorphism of $C$?
What if I do $\langle b_0,b_1,...,b_n,z,z,z, ...\rangle\leftrightarrow \langle b_0,b_1,...,1-b_n,z,z,z ...\rangle$? In other words, looking at $C'$ as the eventually constant sequences in $2^\omega$, and flipping the value of a sequence at last place before it stays constant? That should produce a homeomorphism, but I wonder if it extends...
 A: Your proposed map isn't a homeomorphism, if I understand it correctly - consider the collection of sequences $a_0, a_1, a_2, \cdots$, where $a_i$ is the sequence consisting of $i$ ones followed by constant zeroes. This sequence of sequences has limit $\langle 1, 1, 1, \ldots\rangle$. Your map fixes the sequence, taking $a_i$ to $a_{i - 1}$ - but it takes the limit to $\langle 0, 1, 1, \ldots\rangle$. So this map isn't continuous, let alone a homeomorphism.
Side note: There are related homeomorphisms of $C'$, which do extend to homeomorphisms of $C$. Let $f_n$ be the map which simply flips the $n$th bit of the input. Each $f_n$ is now a homeomorphism (of either $C'$ or $C$, whichever you care more about).
As to whether every homeomorphism of $C'$ extends to a homeomorphism of $C$: yes, because every "irrational" point $r$ in $C$ is a limit of "rational" points $q_n$ from $C'$. Let $h$ be a homeomorphism from $C'$ to $C'$; it takes every clopen subset of $C'$ to a clopen subset of $C'$. The images of the intervals around each $q_n$ that encompass $r$ form a descending sequence of clopen sets around the $h(q_n)$. Since $C$ is compact, there is a (unique) element of $C$ that lies in the closures of all of these sets in $C$ - this can be taken to be $h(r)$. The resulting extension of $h$ can be easily seen to be a homeomorphism.
A: These kind of problems have been well-studied: see this paper's introduction.
The Cantor set is Polish and zero-dimensional and homogeneous. 
This implies it is SLH (strongly locally homogeneous): there is a base of sets $U$ such that for all $x,y \in U$ there is a homeomorphism $f$ of $X$ such that $f(x)=y$ and $f(z) = z$ for all $z \in X\setminus U$.
Another theorem is that LCH Polish spaces are countable dense homogeneous: if $D$ and $E$ are countable dense in $X$ then there is an homeomorphism $h$ of $X$ such that $h[D] = E$. This is not the same as extending a homeomorphism of the dense subsets though. But I think the SLH property combined with the inductive convergence criterion allow you to do this for SLH Polish spaces, including the Cantor set.
A: There are easy examples of  homeomorphisms that cannot be extended to $C$.
For instance, take $p \in (C\setminus C') \cap (0, \frac12)$ and define 
$$h(x) = \cases{ 1 - x & if $p < x < 1-p$ \\ x & otherwise.}
$$
This is a continuous function on $C'$ because $C' \cap (p, 1-p)$ is clopen in $C'$. It also is its own inverse, so it is a homeomorphism.
Since $p$ is not a boundary point of a deleted interval, it is a limit point of both $C' \cap (0, p)$ and $C' \cap (p, 1-p)$. Therefore the oscillation of $h$ at $p$ is $1-2p$, which is positive.
It follows that $h$ cannot be
 extended to a continuous function on $C$.
