Is it true that $(X,T^k)$ minimal for all $k\geq1$ implies $\mathrm{Aut}(X,T) = \mathrm{Aut}(X,T^k)$ for all $k\geq1$? Let $(X,T)$ be a topological dynamical system ($X$ is compact metric space and $T\colon X\to X$ a homeomorphism). Recall that its automorphism group is
$$ \mathrm{Aut}(X,T) = \{g\colon X\to X : \text{$g$ is hoemomorphism and $g\circ T = T\circ g$}\}.$$
Observe that when $(X,T^k)$ is not minimal, we can decompose $X = X_1 \cup X_2 \cup \dots \cup X_r$, where $r\geq2$ and each $X_i$ is $T^k$-invariant, and thus any function $g\colon X\to X$ of the form $g(x) = T^{\sigma(i)}x$ for $x \in X_i$ and $i=1,\dots,r$, with $\sigma\colon\{1,\dots,r\}\to\{1,\dots,r\}$ a bijection, belongs to $\mathrm{Aut}(X,T^k)$. Then, a careful choice of $\sigma$ gives us that $\mathrm{Aut}(X,T) \subsetneq \mathrm{Aut}(X,T^k)$.
My question is about the converse: if $(X,T^k)$ is minimal for any $k\geq1$ then, is it true that $\mathrm{Aut}(X,T) =\mathrm{Aut}(X,T^k)$ for all $k\geq1$?
 A: Here's a subshift counterexample. Let $E : \{0,1\}^* \to \{0,1\}^*$ be the map on finite words that flips every second bit, preserving word length, e.g. $E(01000) = 11101$, and let $O$ flip the even positions. This gives an action of the four-group $V = \langle E, O \rangle \cong (\mathbb{Z}/2\mathbb{Z})^2$ on binary words. Let $u_0, v_0$ be binary words of coprime lengths, say $u_0 = 00, v_0 = 000$. Now, inductively construct $u_{n+1}, v_{n+1}$ by concatenating the words $u_n,v_n$ so that

*

*$|u_{n+1}|$ and $|v_{n+1}|$ are coprime,

*$F(u_n)$ and $F(v_n)$ appear in $F'(u_{n+1})$ and $F'(v_{n+1})$ in all positions modulo $n!$, for all $F,F' \in \langle E,O \rangle$.

It is easy to enforce these, just concatenate words at random to force 2), and then use very advanced number theory to get 1). Now let $X \subset \{0,1\}^{\mathbb{Z}}$ be the subshift of all bi-infinite words $\{0,1\}^{\mathbb{Z}}$ whose finite subwords appear in some $u_n$.
Now, $X$ is totally minimal, i.e. every power of the shift is minimal: consider a power $m$ and a word $w$ which appears in $X$. Then $w$ appears in some $u_n$ where we may assume $n \geq m$, and thus $w$ appears in all positions modulo $n!$ (thus also modulo $m$) in all the words in $V \cdot \{u_{n+1}, v_{n+1}\}$. Every point in $X$ is a concatenation of such words, and we hit a copy of $w$ at least once when jumping over any of them, no matter what the phase, thus we hit $w$ with bounded gaps. Since $E(u_n)$ is in the language of $X$, the language of $X$ is closed under $E$. We thus have $E \in \mathrm{Aut}((X, \sigma^2))$, but obviously $E \notin \mathrm{Aut}((X, \sigma))$. This concludes the proof.
Some possible improvements: I think you can make this into a substitution, by using a variant of the substitutive construction in Example 1 in [1] which is not constant-length, and then check that your example is topological weak mixing (I think it's decidable, or at least we know some necessary and sufficient conditions), which implies total minimality.
Alternatively, by making the choices in a very strange way, you can kill all undesired elements of the automorphism groups of $(X,\sigma)$ and $(X,\sigma^2)$ to get $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z} \times V$ as automorphism groups, respectively. So the groups are not even isomorphic. Probably the best way to prove such a thing is via asymptotic pairs as in [2,3,4].
[1] Salo, Ville, Toeplitz subshift whose automorphism group is not finitely generated, Colloq. Math. 146, No. 1, 53-76 (2017). ZBL1377.37021.
[2] Donoso, Sebastián; Durand, Fabien; Maass, Alejandro; Petite, Samuel, On automorphism groups of low complexity subshifts, Ergodic Theory Dyn. Syst. 36, No. 1, 64-95 (2016). ZBL1354.37024.
[3] Cyr, Van; Kra, Bryna, The automorphism group of a shift of linear growth: beyond transitivity, Forum Math. Sigma 3, Paper No. e5, 27 p. (2015). ZBL1321.37010.
[4] Coven, Ethan; Quas, Anthony; Yassawi, Reem, Computing automorphism groups of shifts using atypical equivalence classes, Discrete Anal. 2016, Paper No. 3, 28 p. (2016). ZBL1378.54035.
