What I wanted to say is that the first and second paragraph are not equivalent. Thus, I took the freedom to answer the question I can :) Anyway, the other question about the dependency of the moments is indeed interesting and perhaps there's something buried in the Weingarten literature.This is not clear to me, and I couldn't find anything useful by looking in recent reviews (e.g. Collins 2022). I'll think about it. (for now I updated my answer for clarification)

No, you are not missing something. We have tried to prove similar statement for filtered randomized benchmarking arxiv.org/abs/2212.06181 (which can be seen as a generalization of XEB), and one can generally construct adversarial noise which increases the XEB. Similar things are true for shadows btw arxiv.org/abs/2310.19947. What you read is the intuitive expectation, which you can prove for local, gate-independent noise (but on average). A proof for more general noise models is however missing to the best of my knowledge.

@user290109 Yes, indeed.

@user290109 the statement is about norms on different vector spaces: we have that $\| (A\otimes\mathbb{1})|\psi\rangle\langle\psi|(A\otimes\mathbb{1})^\dagger\|_1 = \| (A\otimes\mathbb{1})\psi\|_{\ell_2}^2$ because we're computing the trace norm of a rank one operator. Maximizing the latter expression over $\psi$ yields the operator norm $\|A\otimes\mathbb{1}\|_\infty^2=\| A \|_\infty^2$.

I am aware that you formulated the question for a single qubit. Still, the expression for the unital channel $T$, which you give in the 2nd paragraph, is not correct.

Not sure if I get this right. The automorphisms of the generators are still the same, but if you look at a different representation (with higher dimension), then this does not define a unique unitary anymore (not even up to phase), but rather a equivalence class (as "Paulis" are not a basis anymore). Thus, it is highly unclear how efficient classical simulation should work in the context. There are higher-dimensional Cliffords, but they do not come from su(2) representations quantumcomputing.stackexchange.com/questions/35297/…

Is there any reason why this claim should be true? The arguments you put forward do not hold: Only some unital qubit channels are diagonal in the Pauli basis. Just take any unitary. It acts as a $SO(3)$ matrix on the traceless subspace. In particular, it is not necessarily diagonal. Even if you take off-diagonal elements into account, the traceless part of a unital channel is not an SO matrix in general. Although unital channels are in general not even convex combinations of unitaries, this is true for a single qubit (see Watrous Thm. 4.23).

@BenFoxman no, it does not. If they would have the same commutation relations as the standard basis, the basis change matrices would be symplectic. But they can in fact be arbitrary, and a brief calculation shows that the combination of both basis changes, i.e. $S$, is symplectic, so it corresponds to a Clifford unitary (up to Paulis). More generally, if you have a symplectic matrix $L$, you can represent it in some other basis as $M_B L M_A^{-1}$, see e.g. en.wikipedia.org/wiki/…

@CraigGidney sure, but this was an assumption in Ben's question.