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Sep 27, 2022 at 15:51 comment added YCor Just using plain conjugation you get exponentially many. More precisely, if the rank is $d$, the $(r/2)$-ball has about $(2d-1)^{r/2}$ elements and hence you get about $(2d-1)^{r/2}$ such subsets in the $r$-ball. On the other hand, a trivial upper bound is the number of $d$-tuples in the $r$-ball, and this is $(2d-1)^{dr}$. It remains to see the exact exponential rate, which is thus between $(1/2)\log(2d-1)$ and $d\log(2d-1)$.
Sep 27, 2022 at 15:51 comment added Denis T There's a somewhat relevant folklore result which states that if an automorphism $f$ of a free group takes only small (relative to $n$) portion of an $n$-ball outside of it (in a paper by Shpilrain or some of his usual co-authors I remember seeing effective bounds; having that portion tend to zero is definitely enough) then $f$ lies in signed symmetric group acting on generators.
Sep 27, 2022 at 15:20 history asked 24601 CC BY-SA 4.0