Question: I would like to find two topological dynamical systems that are not topologically conjugate but nevertheless have the same topological entropy.

Two topological dynamical systems $f:X\to X,g:Y\to Y$ are topologically conjugate if there is a homeomorphism $H:X\to Y$ such that $f\circ H=H\circ g$.

If $X$ is a compact metric space and $f:X\to X$ is continuous, then

$d_n(x,y):=\max\{d(f^k(x),f^k(y))|0\leq k\leq n-1\}$;

$\forall \varepsilon>0, N(n,\varepsilon):=\max\{|\{p_1,...,p_m\}|=m|i\neq j \implies d_n(p_i,p_j)\geq\varepsilon\}$

(i.e., $N(n,\varepsilon)$ is the largest number of points $p_1,...,p_m\in X$ such that $i\neq j\implies d_n(p_i,p_j)\geq\varepsilon$);

and $h(f):=\lim_{\varepsilon\to0}\limsup_{n\to\infty}\dfrac{1}{n}N(n,\varepsilon)$ is the topological entropy of $f$ (This is the definition of Bowen and Dinaburg, and it requires a metric. For a definition that works for the general setting see wikipedia).

Motivation: We know that topological conjugacy preserves topological entropy. But the converse is not true: first example that comes to mind is that $h(\sigma^+_k)=\log k=h(\sigma_k)$ but $\sigma^+_k\not\sim \sigma_k$, where $\sigma^+_k:\Sigma^+_k\to \Sigma^+_k$ is the topological Bernoulli shift and $\sigma_k:\Sigma_k\to \Sigma_k$ is the full shift. But it is immediate that these two systems are not conjugate (full shift is bijective, the other is not). So I am more interested in finding a counterexample where both of the systems are invertible, though any other counterexamples, either that use metric spaces or topological spaces, are welcomed.

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    $\begingroup$ Take $f,g : [-1,1] \to [-1,1]$ defined by $f(x)=x$, $g(x)=-x$. Both have entropy zero. $\endgroup$ – Lee Mosher Mar 16 '15 at 23:13

Let $R_\alpha$ an irrational rotation of the $S^1$ and $R_\beta$ a rational rotation of the $S^1$. Then, $h_{top}(R_\alpha)=h_{top}(R_\beta)=0$, but are not topologically conjugate.


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