Does every homeomorphism of a compact metric space lift to the Cantor set? This is a follow-up to this question.  It is well-known that any compact metrizable space can be expressed as a quotient of the Cantor set.  But can every homeomorphism of such a space be lifted to a homeomorphism of the Cantor set?
More precisely, let $K$ be a compact metric space, and let $h\colon K\to K$ be a homeomorphism.  Does there exist a quotient map $q\colon C\to K$, where $C$ is the Cantor set, and a homeomorphism $\bar{h}\colon C\to C$ making the following diagram commute?
$$
\begin{array}{ccl}
C & \xrightarrow{\textstyle\;\;\;\bar{h}\;\;\;} & C \\[8pt]
q\left\downarrow\begin{array}{c}\! \\\! \end{array}\right. & & \left\downarrow\begin{array}{c}\! \\\! \end{array}\right.\!\!q \\[8pt]
K & \xrightarrow[\textstyle\;\;\;h\;\;\;]{} & K
\end{array}
$$
 A: This is true for actions of countable groups, and in particular for $\mathbb{Z}$-actions.
Let $C$ be a Cantor set and $G$ a countable group. Recall that $C^G$ is also a Cantor set. We let $\lambda:G\curvearrowright C^G$ denote the usual left action: $\lambda_g(f)(h)=f(g^{-1}h)$ for all $g,h\in G$ and $f\in C^G$.
Now, suppose that $G$ acts freely by homeomorphisms on a compact metric space $K$. Let $q:C\to K$ be a quotient map. Define the new space
\begin{align*}
D&=\left\{f\in C^G:\forall g\in G, q(f(g))=g^{-1}(q(f(1)))\right\}\\
&=\left\{f\in C^G:\forall g,h\in G, q(f(hg))=g^{-1}(q(f(h)))\right\}
\end{align*}
At this step, $D$ does not need to be a Cantor set, however it should be clear that $D$ is closed in $C^G$, hence compact, metrizable, zero-dimensional. The only point that might fail is perfectness, but we will deal with this later on.
Note that $D$ is $\lambda$-invariant, i.e., if $f\in D$ and $g\in G$ then $\lambda_g f\in D$.
Now, let $Q:D\to K$ be "evaluation at $1$ followed by $q$": $Q(f)=q(f(1))$. One easily checks that this is surjective, hence a quotient map, which is $(\lambda,G)$-invariant, i.e., $Q\lambda_g=gQ$ for all $g\in G$.
$$
\begin{array}{ccl}
D & \xrightarrow{\textstyle\;\;\;\lambda_g\;\;\;} & D \\[8pt]
Q\left\downarrow\begin{array}{c}\! \\\! \end{array}\right. & & \left\downarrow\begin{array}{c}\! \\\! \end{array}\right.\!\!Q \\[8pt]
K & \xrightarrow[\textstyle\;\;\;g\;\;\;]{} & K
\end{array}
$$
We would be done if $D$ was already perfect, but this might not me the case. To fix this, we simply consider $D'=D\times C$, which is a Cantor set, and extend the action of $G$ and the quotient map from $D$ naturally to $D'$, namely via $\lambda_g'(f,x)=(\lambda_gf,x)$ and $Q'(f,x)=Q(f)$ for all $(f,x)\in D'$. Then $D',\lambda'$ and $Q'$ have the properties we want.

We can improve our previous result and show that (infinite) minimal systems are factors of Cantor minimal systems. By a compact dynamical system we mean a compact metric space $K$ endowed with an action of a countable group $G$ by homeomorphisms. A subsystem consists of a closed invariant subset $M$ of $K$, endowed with the restricted action. A (sub)system is called minimal if it does not have any nontrivial subsystems. A Cantor system is a compact dynamical system on the Cantor set.
Lemma: Suppose $(C,G)$ is a Cantor system and $M$ is a minimal (closed) subsystem. Then $M$ is either finite or a Cantor set.
Proof: The only problem is to show that $M$ is perfect. Since $M$ is a minimal subsystem, it is the closure of the orbit of any of its points. Suppose $M$ is not finite and let $x\in M$. Since $M=\overline{Gx}$, the orbit $Gx$ must be infinite and hence accumulates at some point, i.e., there exist a sequence of elements $g_n\in G$ such that $g_nx\neq g_mx$ for $n\neq m$, and $g_nx\to y$ for some $y\in M$. Let's analyse two cases separately:
Case 1: $y\in Gx$.
Then $y=hx$ for some $h$, so $\left\{h^{-1}g_nx\right\}$ is a sequence of distinct points which converges to $x$.
Case 2: $y\not\in Gx$, or equivalently $x\not\in Gy$.
Since $M=\overline{Gy}$, for any neighbourhood $U$ of $x$ there exists $h$ such that $hy\in x$, and $hy\neq x$ by hypothesis. QED
Theorem: Any infinite minimal compact dynamical system $(K,G)$ is a factor of a minimal Cantor system.
Proof: We already know that $(K,G)$ is a factor of a Cantor system, say $Q:C\to K$ is a quotient map. Let $(D,G)$ be a minimal subsystem of $(C,G)$. Since $Q$ maps orbits of $(D,G)$ to orbits in $(K,G)$, and the latter is minimal, the restriction $Q|_D:D\to K$ is surjective, so in particular $D$ cannot be finite, hence it is a Cantor set. In other words, $Q|_D$ is a quotient map from the Cantor minimal system $(D,G|_D)$ onto $(K,G)$. QED
