Suppose that a bounded sequence of real numbers $s\_i$ ($i\in\omega$) has a limit $\alpha$ along some ultrafilter $\mu\_1\in \beta{\Bbb N}\setminus{\Bbb N}$. Then given another ultrafilter $\mu\_2\in \beta{\Bbb N}\setminus{\Bbb N}$, surely there exists some rearrangement $s\_{r(i)}$ of $s\_i$ that has the same limit $\alpha$.


One can easily extend this simple observation to a countable family of sequences. 


Now given $s\_{i;j}$ ( $i,j\in \omega$; values bounded for each fixed $j$) with limits $\alpha\_j$ along a fixed $\mu\_1\in \beta{\Bbb N}\setminus{\Bbb N}$, and given another ultrafilter $\mu\_2\in \beta{\Bbb N}\setminus{\Bbb N}$, there exists a *simultaneous* rearrangement $s\_{r(i);j}$ having the same limits $\alpha\_j$ along $\mu\_2$. 


All this fails if we pass to size $c=2^\omega$ families of sequences. Indeed $s\_{i;j}$ could then enumerate all bounded sequences. But all the limits $\alpha\_j$ together would determine $\mu\_1$. Taking limits of a simultaneous rearrangement of all the sequences amounts, equivalently, to taking limits of the original sequences along an ultrafilter $\mu\_2'$ in the orbit of $\mu\_2$ under the action of the symmetric group of $\Bbb N$ extended to $\beta\Bbb N$. Equality of all those limits thus forces $\mu\_1=\mu\_2'$, and that places $\mu\_1$ and $\mu\_2$ in the same orbit of the symmetric group action, a severe restriction on $\mu\_2$.


**Question**: If CH fails, what happens for a size $\omega\_1$ family of sequences?



Limits of rearranged sequences along ultrafilters
If $\mathfrak{p} > \kappa$, then you can get such a $\pi$ for every family of $\kappa$-many bounded sequences.


To see this, suppose $\mathfrak{p} > \kappa$, $\mu\_1$ and $\mu\_2$ are nonprincipal ultrafilters, and $s^\xi\in\mathbb{R}^\mathbb{N}$ ($\xi < \kappa$) is a family of bounded sequences. Let $\lambda^\xi$ be the $\mu\_1$-limit of $s^\xi$ and let $$ X\_n^\xi = \left\{m\in\mathbb{N}\; :\; |s\_m^\xi - \lambda^\xi| < \frac{1}{n}\right\} $$ Then each $X\_n^\xi$ is a member of $\mu\_1$, and hence every intersection of finitely-many of them is infinite. Since $\mathfrak{p} > \kappa$, there is then an infinite set $P$ which is contained (mod-finite) in all of them. Let $\pi$ be any permutation such that $\pi''(P)\in \mu\_2$. Then if $t^\xi$ is the rearrangement of $s^\xi$ by $\pi$, we have $t^\xi\_m = s^\xi\_{\pi^{-1}(m)}$, so


$$ Y^\xi\_n = \left\{ m\in\mathbb{N}\; :\; |t\_m^\xi - \lambda^\xi| < \frac{1}{n}\right\} = \pi''(X^\xi\_n) \supseteq^\* \pi''(P) $$


and hence all of these $Y^\xi\_n$'s are in $\mu\_2$, which means that the $\mu\_2$-limit of each $t^\xi$ is still $\lambda^\xi$.


It would be nice to improve this to a larger cardinal characteristic than $\mathfrak{p}$.


EDIT: I think $\mathfrak{u}$ (the minimal size of a base for a nonprincipal ultrafilter) is an upper bound.


Let $\mu\_1$ be a nonprincipal ultrafilter with a base $\mathcal{B}$ of minimal size. Consider the family of sequences $\chi\_A$ ($A\in\mathcal{B}$), where $\chi\_A$ is the characteristic function of $A$. The limit of each such $\chi\_A$ along $\mu\_1$ is $1$. If $\pi$ is a permutation, then the sequence we get by rearranging $\chi\_A$ using $\pi$ is exactly $\chi\_{\pi^{-1}(A)}$. So, given $\mu\_2$, we'd like to find a $\pi$ such that for each $A\in\mathcal{B}$, $\pi^{-1}(A)\in \mu\_2$. Clearly this implies that $\pi^{-1}(A)\in \mu\_2$ for all $A\in \mu\_1$. But this implies that $\mu\_2$ is exactly


$$ \{\pi^{-1}(A) \; : \; A\in\mu\_1\} $$


(If there were some $B\in\mu\_2$ not in the above set, then consider $\pi''(B)$. This is not in $\mu\_1$ by assumption, hence $\pi''(\mathbb{N}\setminus B) \in \mu\_1$, hence $\mathbb{N}\setminus B\in \mu\_2$, a contradiction.)


So in this case we can only find such a $\pi$ when $\mu\_2$ is in the orbit of $\mu\_1$ under $S\_\mathbb{N}$, and of course this orbit is not all of $\beta\mathbb{N}\setminus \mathbb{N}$. (There are only continuum-many points in the orbit of $\mu\_1$ but $2^{\mathfrak{c}}$-many ultrafilters total.)


