Are these permutation groups, defined by asymptotic properties, isomorphic?
(I always find depressing to see group theory classified as "abstract" algebra at some places.)
It's correct they're pairwise non-isomorphic (namely, the groups $H_r=\{\sigma\in S_\mathbf{N}: c_\sigma(n)=o(n^r)\}$ for $r\in \mathopen]0,1]$, denoted $G_{x^r}$ in the question).
The first observation is that they're not conjugate: this is easy (using that for every permutation $f$ of $\mathbf{N}$ there's an infinite subset $I$ such that $f(n)\ge n$ for all $n\in I$.
Edit this non-conjugation claim is unclear at this point. So for the moment this post reduces showing that they're non-conjugate, and it turn this means: for $0<r<s\le 1$, does there exist a permutation $f$ of $\mathbf{N}$ such that $f(N_r)=N_s$, where $N_u$ is the set of subsets $I$ of $\mathbf{N}$ that are $n^u$-negligible: $\#(N_u\cap [n])=o(n^u)$
Next, we see that an isomorphism between two such groups would be implemented by a conjugation. Namely we have the following lemma (which is probably a very particular result of M. Rubin's theorems):
Lemma let $X$ be an infinite set; let $U,V$ be subgroups of $S_X$ that contain the group $S_X^\#$ of finitely supported permutations of $X$. Let $f$ be an isomorphism $U\to V$. Then $f$ is the restriction of a conjugation of $S_X$: there exists $\alpha\in S_X$ such that $f(g)=\alpha g\alpha^{-1}$ for all $g\in U$. In particular, $V=\alpha U\alpha^{-1}$.
To prove the lemma, the first step is to check that $f(S_X^\#)=S_X^\#$: namely we need to "recognize" $S_X^\#$ within both $U$ and $V$. This is done in "Lemma*" below.
The second step is a classical one: $\mathrm{Aut}(S_X^\#)=S_X$ (acting by conjugation). This can be found in the books by Scott or Dixon-Mortimer. So, there exists $\alpha\in S_X$ such that $f(w)=\alpha w\alpha^{-1}$ for all $w\in S_X^\#$.
The third and concluding step: for $g\in U$ and $w\in S_X^\#$, we compute $f(gwg^{-1})$ in two ways: first it equals $f(g)f(w)f(g)^{-1}=f(g)\alpha w\alpha^{-1}f(g)^{-1}$, second it equals $\alpha gwg^{-1}\alpha^{-1}$. Since the centralizer of $S_X^\#$ is trivial, we deduce $\alpha g=f(g)\alpha$ for all $g\in U$, which is the desired form.
Lemma$^*$ Let $U$ be a subgroup of $S_X$ containing $S_X^\#$. Let $T_U$ be the set of elements $s$ of order 2 in $U$ such that for every $U$-conjugate $t$ of $s$, either $t$ commutes with $s$ or $(st)^3=1$. Then $T_U$ is the set of transpositions of $X$. In particular, $\langle T_U\rangle=S_X^\#$ is a characteristic subgroup of $U$, and for any two such groups $U_1,U_2$, every isomorphism $U_1\to U_2$ induces an automorphism of $S_X^\sharp$.
It is clear that a transposition satisfies this property. Conversely, suppose that $s$ has order 2 and is not a transposition. Say, $s(a)=b$ and $s(c)=d$ with $|\{a,b,c,d\}|=4$.
First case, $s$ fixes at least two points. Then there exists a $S_X^\#$-conjugate $t$ of $s$ fixing $a,d$ and exchanging $b,c$. Then $st$ has order 4 on $\{a,b,c,d\}$.
Second case, $s$ fixes at most one point. Then there exist $e,f,g,h$ such that $s(e)=f$ and $s(g)=h$, and $|\{a,b,c,d,e,f,g,h\}|=8$. Hence there exists a $S_X^\#$-conjugate of $s$ exchanging all pairs $\{b,c\}$, $\{d,e\}$, $\{f,g\}$, $\{h,a\}$. Then $st$ has order 4 on $\{a,b,c,d,e,f,g,h\}$.
In both cases, we see that $t$ and $s$ don't commute and $(st)^3\neq 1$.
A proof that the groups $G_{x^r}$ are not conjugate, which proves that they are not isomorphic by YCor's answer:
Lemma: For any permutation $\alpha : \mathbb{N} \to \mathbb{N}$, and any $t \in (0, 1]$, there is a set $B \subset \mathbb{N}$ with $|B \cap [n]| = O(n^t)$ and $|\alpha(B) \cap [n]| = \Omega(n^t)$.
Proof: For $k \geq 0$, note that the set $A_k = \alpha^{-1}([2^{k+1}]) \setminus [2^k]$ has size at least $2^k$, so let $B_k$ be a subset of $A_k$ of size $2^{kt}$, and define $B = \bigcup_k B_k$. Then for $2^k < n \leq 2^{k+1}$, all $B_j$ with $j \geq k+1$ are disjoint from $[2^{k+1}]$, hence $$B \cap [n] \subset B \cap [2^{k+1}] \subset \bigcup_{j=0}^k B_j$$ which gives the bound $|B \cap [n]| \leq \sum_{j=0}^k 2^{jt} \leq C \cdot 2^{kt} \leq Cn^t$, meaning $|B \cap [n]| = O(n^t)$. In the other direction, for $n \geq 2$ we have $2^k \leq n < 2^{k+1}$ for some $k$, hence since $\alpha(B_{k-1}) \subset \alpha(B) \cap [2^k]$ this gives $$|\alpha(B) \cap [n]| \geq |\alpha(B) \cap [2^k]| \geq |\alpha(B_{k-1})| = 2^{(k-1)t} \geq 2^{(k+1)t - 2} \geq \frac{1}{4} n^t$$ so $|\alpha(B) \cap [n]| = \Omega(n^t)$.
Now, suppose there is an isomorphism $f$ from $G_{x^s}$ to $G_{x^r}$, for some $0 < r < s \leq 1$. By YCor's answer, $f$ takes the form $f(g) = \alpha g \alpha^{-1}$ for some permutation $\alpha : \mathbb{N} \to \mathbb{N}$. Take $t$ with $r < t < s$, and let $B \subset \mathbb{N}$ be as in the Lemma. Now let $\pi$ be a permutation $\mathbb{N} \to \mathbb{N}$ with $\{k \in \mathbb{N} \mid \pi(k) \neq k\} = B$; it is straightforward to construct such a permutation. We have $c_\pi(n) = |B \cap [n]| = O(n^t) = o(n^s)$, so $\pi \in G_{x^s}$, and thus $\sigma := f(\pi)$ is in $G_{x^r}$. But $\sigma = \alpha \pi \alpha^{-1}$, so $\{k \in \mathbb{N} \mid k \neq \sigma(k)\} = \{\alpha(k) \mid k \neq \pi(k)\} = \alpha(B)$, so $c_\sigma(n) = |\alpha(B) \cap [n]| = \Omega(n^t)$, which contradicts $\sigma \in G_{x^r}$. Thus $G_{x^r}$ and $G_{x^s}$ are not isomorphic.