*This question came out of a conversation with my students about Riemann's rearrangement theorem, and the general problem of which permutations are "safe" w/r/t summing infinite series.*

Let $S_\infty$ be the group of permutations of $\mathbb{N}$. For a sequence $\mathscr{A}=(a_i)_{i\in\mathbb{N}}$, say that a permutation $p\in S_\infty$ is **$\mathscr{A}$-placid** iff for every $q\in S_\infty$ and every pair of integers $z_0, z_1$ we have $$\sum_{i\in\mathbb{N}}a_{q(i)}\simeq \sum_{i\in\mathbb{N}}a_{p^{z_0}\circ q\circ p^{z_1}(i)}$$ where "$s\simeq t$" means "either $s$ and $t$ are each undefined, or they are defined and equal." Basically, $p$ is $\mathscr{A}$-placid if $p$ is never interesting from the point of view of rearranging the terms in $\mathscr{A}$. For example, the permutation swapping $2i$ and $2i+1$ for each $i$ is $\mathscr{A}$-placid for every $\mathscr{A}$.

I'm curious whether placidity actually depends on the sequence in question *(restricting attention to sequences whose corresponding series converge conditionally, to avoid triviality)*. For example, are the following equivalent?

$p$ is placid with respect to the alternating harmonic sequence $((-1)^{i+1}{1\over i})_{i\in\mathbb{N}}$.

$p$ is $\mathscr{A}$-placid for every conditionally convergent sequence $\mathscr{A}$.

I suspect the answer is negative, but I don't immediately see how to prove it.

EDIT: as far as I can tell, the only "obviously placid" permutations are those in which there is a finite bound on the distance an element of $\mathbb{N}$ is moved. Merely having finite orbits isn't enough: for example, for any conditionally convergent sequence $\mathscr{S}$ there is a permutation of $\mathbb{N}$ of order $2$ which applied to $\mathscr{S}$ results in a series with limit $+\infty$. This was pointed out to me by a colleague after I brashly claimed otherwise!