Let $\{a_n\}_{n\in \mathbb{N}}$ be a sequence of real numbers such that the series $\sum a_n$ is conditionally convergent, i.e. the limit $\lim\limits_{N\to \infty} \sum_{n=0}^Na_n =:L \in \mathbb{R}$ exists but $\sum_{n=0}^\infty \vert a_n \vert = \infty $.
Now by the Riemann rearrangement theorem for every $r \in \mathbb{R}$ there exists a permutation $\sigma \in \mathrm{Aut}(\mathbb{N})$ such that $\lim_{N\to \infty} \sum_{n=0}^N a_{\sigma(n)} = r$. Lets write $S(\sigma) = r$ in this situation.
My (very vague) question is the following:
Is there a method to determine the subset of all permutations $\sigma \in \mathrm{Aut}(\mathbb{N})$ such that $S(\sigma) = L$ (= the original limit of the series). Of course this contains all permutations which only permute a finite subset of the natural numbers, but I am mostly interested in how the sequence $\{a_n\}$ shapes this set.
EDIT: As pointed out by Martin in the comments, there is an answer for the subset of permutations that fix the original limit for all sequences.