I posted [this question](https://math.stackexchange.com/questions/1383730/rearrangements-that-never-change-the-value-of-a-sum) on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$,
$$
\lim_{n\to\infty} \sum_{k=1}^n a_n = \lim_{n\to\infty} \sum_{k=1}^n a_{f(n)},
$$
where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them.  That bijection never changes the values of sums.  It's a countably infinite sequence of binary choices, so it's uncountable.)