Thompson's group $V$, which is a finitely presented infinite simple group, consists of all homeomorphisms of the Cantor set $\{0,1\}^\mathbb N$ that can be described the following way. A *prefix code* is a collection of finite words none of which is a prefix of another. A finite prefix code $C$ is maximal if it is not contained in another prefix code. Equivalently, it is maximal if removing the vertices of the binary tree corresponding to $C$ disconnects it.

An element of Thompson's group is given by specifying two maximal prefix codes $C_1=\{u_1,\ldots, u_n\}$ and $C_2=\{v_1,\ldots, v_n\}$ of the same size and a permutation $\sigma\in S_n$. The action of the corresponding element $f_{C_1,C_2,\sigma}$ on $x\in \{0,1\}^n$ is as follows. There is a unique factorization $x=u_iz$ with $u_i\in C_1$. Then $f_{C_1,C_2,\sigma}(x) = v_{\sigma(i)}z$. Note that the prefix codes $C_1,C_2$ are not uniquely determined by the element of $V$.

Let $X=\{0,1\}^\mathbb N$ and let $R$ be an equivalence relation such that $Y=X/R$ is compact Hausdorff and the projection $X\to Y$ is $V$-equivariant. Suppose that $R$ is not the equality relation and has more than one equivalence class. Note that $R$ must be closed as a subset of $X\times X$ by Hausdorffness. Since the complement of $R$ in $X\times X$ is non-empty and open, it follows that there are finite words $u,v$ such that $ux$ and $vy$ are never equivalent for any infinite words $x,y$. By extending the shorter of the two words, we may assume $|u|=|v|=m$ (where $|\cdot|$ is the length).

Since $R$ is not the equality relation we can find two distinct infinite words $x_1,x_2$ which are equivalent under $R$. I claim we can find an element $g\in V$ such that $g(x_1)\in u\{0,1\}^\mathbb N$ and $g(x_2)\in v\{0,1\}^\mathbb N$. This will contradict that $X\to Y$ is $V$-equivariant.

Let $n$ be the length of the longest common prefix of $x_1,x_2$. Let $k>\max\{n,m\}$ and let $C$ be the maximal prefix code consisting of all words of length $k$. Then $x_1,x_2$ have different prefixes $p,q$ of length $k$. Also, since $k>m$, we can find $p',q'\in C$ such that $u$ is a prefix of $p'$ and $v$ is a prefix of $q'$. There is an element $g=f_{C,C,\sigma}\in V$ such that $g$ takes all infinite words $px$ to $p'x$ and $qx$ to $q'x$. It follows that $g(x_1)$ and $g(x_2)$ are not equivalent under $R$, a contradiction.