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Given a topological dynamical system $(X,T)$ (so that $T$ is a homeomorphism of the compact metric space $X$) and a point $x\in X$ we call the set ${\mathcal O}(x):=\overline{\{T^nx:n\in\mathbb Z\}}$ the orbit closure of $x$.

Question 0: Is there a name for systems with the property that the orbit closure of every point is uniquely ergodic (i.e., supports a unique invariant measure)?

It is well known that nilsystems have this property but not all distal systems do.

Question 1: Is it true that if $(X,T)$ and $(Y,S)$ are uniquely ergodic, then $(X\times Y,T\times S)$ has the property that every orbit closure is uniquely ergodic?

Question 2: What if in addition $(X,T)$ and $(Y,S)$ are distal?

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I think the two-sided Morse system is a counterexample to both questions the first question. Consider the two points $x^{(1)}=\ldots 0110100110010110\cdot0110100110010110\ldots$ and $x^{(2)}=\ldots 1001011001101001\cdot0110100110010110$. The orbit closure of $(x^{(1)},x^{(2)})$ supports exactly two ergodic invariant measures, the diagonal measure on the Morse system and the anti-diagonal measure.

By the way, I have a paper with Emmanuel Lesigne and Máté Wierdl establishing lots of strange properties of the Morse system (such as for any point, $x=(x^{(1)},x^{(2)},x^{(3)})$ of $M^3$, $x$ is generic for some ergodic invariant measure, but there exist points of $M^4$ that are not generic for any invariant measure).

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    $\begingroup$ Thanks for the example and the reference. This does answer the first question, but I'm not sure the Morse system is distal. In fact I thought no symbolic system can be distal (unless it is finite). But perhaps one can project onto the largest distal factor and still have the same property? $\endgroup$
    – Joel Moreira
    Commented Jul 11, 2017 at 9:45
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    $\begingroup$ Oops. I'm not so used to thinking about distal systems. The lack of distality is shown exactly by the two points that I use (and essentially no other points). I suspect the maximal distal factor is just the odometer, which, of course, does have the property you are asking about. $\endgroup$
    – Anthony Quas
    Commented Jul 11, 2017 at 14:17
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Not full answer, but too long for a comment.

The paper The structure of mean equicontinuous group actions defines your property under the name "pointwise unique ergodicity." Their Theorem 1.2 doesn't quite answer your question, but it's interesting: it says that $(X, T)$ has a property called "mean equicontinuity" if and only if the product system $(X \times X, T \times T)$ is pointwise uniquely ergodic AND the function $(x, y) \mapsto \mu_{(x, y)}$ is weak-* continuous.

In addition, Corollary 3.6(2) from Mean equicontinuity and mean sensitivity states that if $(X,T)$ is distal and mean equicontinuous, then it is equicontinuous.

This shows that if $(X, T)$ is distal and not equicontinuous, then even if the product system is pointwise uniquely ergodic, the function assigning measures to points isn't weak-* continuous. I guess even for (non-rotation) nilsystems this is true, since those won't be equicontinuous.

Of course none of this answers your question. Somehow my impression is that there should exist distal systems "so far from equicontinuous" that the product system isn't pointwise uniquely ergodic, but I don't have an example.

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