For simplicity, let $t=1$, let $U$ be uniform on $[0,1]$ and independent of $(Z_s)$, and let $(Y_s)$ be the process you get by shifting by $U$. Then $Y_1 = Z_1$ and $(Y_s)$ has the same distribution as $(Z_s)$. Conditional upon $Z_1$ (ie on $Y_1$) $(Y_s)$ is distribution as a uniformly random cyclic shift of conditioned $(Z_s)$, so conditioned $(Y_s)$ has cyclically exchangeable increments. But conditioned $(Y_s)$ and conditioned $(Z_s)$ have the same distribution. This type of argument goes back to Sparre Andersen and is very useful for proving Ballot-theorem-style results.

For my last paragraph I should have said that all that holds conditional upon $Z_t$, or else it doesn't address your question.

Incidentally, Kai Lai Chung's "Course in probability theory" has a very careful exposition of characteristic functions and infinitely divisible distributions and in particular does all the required complex-analytic details related to choosing a branch of complex log, etcetera.

I think what I described is a rigorous proof. Write $\chi_s(z) = \mathbb{E}(e^{izX_s})$. By infinite divisibility of $X_t$, for rational $q \in (0,1)$ and $s=qt$, $\chi_s(z)=(\chi_t(z))^{q}$ so $\mathbb{E}(X_{qt})=0$. By bounded convergence, the same equality follows for real $r \in (0,1)$. It then follows immediately that $\mathbb{E}(X_{rt}) = \lim_{q \to r} \mathbb{E}(X_{qt}) = 0$. This works for either Ori's or my argument.

I see. The point is that the matching rules in the Penrose tiling precisely imply that the set of possible animals is equal to the set of animals that actually appear in the Penrose tiling. I was missing this point. Thanks!

Thanks for these comments. I meant to restrict the tiling to be aperiodic originally; I've modified the question accordingly.

Thanks Alex. The point of my question is to have a consistent notation for both the former and the latter, do you have a thought about what one might use in this case?