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It is already known that the Clifford hierarchy is not closed under arbitrary products, see this post which shows that the product $ THT $ is not in any level of the hierarchy.

What about products of a gate with itself? If $ g $ is a gate in the hierarchy then are all powers $ g^m $ in the hierarchy as well?

Certainly this is true for all $ k=1,2 $ since these are groups. Also any power of a diagonal gate in the Clifford hierarchy is also in the Clifford hierarchy since diagonal gates in $ \mathcal{C}^{(k)} $ form a group and thus are closed under powers.

A counter example might be something like a gate $ g \in \mathcal{C}^{(3)} $ such that $ g^2 $ is not in $\mathcal{C}^{(3)}$.

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TL;DR: Clifford hierarchy is not closed under raising to integer powers. One suitable counterexample is $g:=TH$ which resides in the third level $\mathcal{C}^{(3)}$ of the Clifford hierarchy, but whose square $g^2=THTH$ is altogether outside the hierarchy $\mathcal{C}=\cup_{i=1}^\infty\mathcal{C}^{(i)}$.

Right-multiplication by Cliffords fixes high levels of the hierarchy

The first level $\mathcal{C}^{(1)}$ of the $n$-qubit Clifford hierarchy is the $n$-qubit Pauli group and the $k$th level is defined recursively as $$ \mathcal{C}^{(k)}:=\{U\in U(2^n)\,|\,\forall P\in\mathcal{C}^{(1)}\quad UPU^\dagger\in\mathcal{C}^{(k-1)}\}.\tag1 $$ In particular, conjugation by any Clifford $C\in\mathcal{C}^{(2)}$ sends any Pauli operator $P\in\mathcal{C}^{(1)}$ to a Pauli operator $Q\in\mathcal{C}^{(1)}$ $$ CPC^\dagger=Q\in\mathcal{C}^{(1)}.\tag2 $$ Similarly, every $U\in\mathcal{C}^{(k)}$ for $k\geqslant 2$ sends every Pauli operator $Q\in\mathcal{C}^{(1)}$ to a unitary in $\mathcal{C}^{(k-1)}$ $$ UQU^\dagger=V\in\mathcal{C}^{(k-1)}.\tag3 $$ Combining $(2)$ and $(3)$, we see that for every Clifford $C\in\mathcal{C}^{(2)}$ and every $U\in\mathcal{C}^{(k)}$ $$ UCP(UC)^\dagger=UCPC^\dagger U^\dagger=UQU^\dagger=V\in\mathcal{C}^{(k-1)}.\tag4 $$ Thus, we obtain the equivalence $$U\in\mathcal{C}^{(k)}\iff UC\in\mathcal{C}^{(k)}\tag5$$ for every Clifford $C\in\mathcal{C}^{(2)}$ and any integer $k\geqslant 2$.

Third level is not closed under squares

Finally, $T\in\mathcal{C}^{(3)}$, so by $(5)$ we have $g:=TH\in\mathcal{C}^{(3)}$. But, we know that $THT$ is not in the Clifford hierarchy $\mathcal{C}$, so using $(5)$ again we obtain$^1$ that $g^2=THTH\notin\mathcal{C}$. Thus, $g$ is an operator in the third level of the Clifford hierarchy whose square is not in any level of the hierarchy.


$^1$ The equivalence in $(5)$ only implies that $g^2$ is not in the second or higher level of the Clifford hierarchy, but $THTH$ is a $2\times 2$ matrix without zero elements, so it is not in $\mathcal{C}^{(1)}$, either.

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