While doing some simulations dealing with the triadic odometer $T$, I observed that the results were the same when I replace $T$ with $T^2$. Therefore I'm wondering whether $T$ is isomorphic to $T^2$. Is it true?



Is the triadic odometer isomorphic to its square?
Yes. More generally, if $T$ is a $k$-odometer, and $r$ is relatively prime to $k$, then $T^r$ is conjugate (that is, measure-theoretically isomorphic) to $T$. Relative primeness is necessary in order that $T^r$ be ergodic. 


One way to prove this (among several) is to use the characterization of odometers by their spectral properties. Another is to go directly to the adding machine, and notice that $T^r$ represents adding $r$, and then drawing the corresponding diagram (I haven't a clue about how to put a commuting diagram here) to show (topological, with this method) conjugacy. 


In more detail (for the latter method), let $A$ be the $k$-adics. Viewed as a ring, $A$ has $r$ as a unit; hence multiplication by $r$, $p:= \times r: A\to A$, is a homeomorphism (continuous, invertible, measure-preserving, everything, etc). Then $pT =T^r p$, yielding a conjugacy. 


And yet another way to see this, is to realize $T$ and $T^r$ as matrix-valued walks, and there is a not-so-difficult argument to show their Poisson boundaries are equal as $Z$-spaces. [This refers to the classification result of Giordano and me, *Matrix-valued random walks and variations on property AT*, Münster J Math 1 (2008) 15-72. This paper is too much to go through merely to obtain this result; however, it can be used to also deal with the case of other odometers, corresponding to supernatural numbers. $k$-odometers form a very small subclass of these.] 


