Is a subgroup determined by where the generators are in its cosets? Let $G$ be a finitely generated infinite group and $H$ be a subgroup of finite index. In particular say $G=\langle x_1,x_2,...,x_n\rangle$ and $G:H=\{R_1,...,R_m\}$. Does the distribution of the $x_i$ among the $R_j$ determine $H$ (up to reordering cosets)?
For example if $G=\langle x_1,x_2,x_3\rangle$ and $H$ is such that $x_1, x_2\in R^H_1$ and $x_3\in R^H_2$, then if $K\le G$ with $x_1,x_2\in R_1^K, x_3\in R_2^K$, we must have $H=K$?
Hopefully the intention is clear, but let me know in the comments if otherwise

I'll presume what is implicit in your notation that the index of $H$ equals the index of $K$, and is denoted $m$.
I'm not entirely sure what "distribution (up to reordering cosets)" really means, but from your example it seems to mean that $K$ and $H$ induce the same partition of the generating set $\{x_1,x_2,...,x_n\}$.
One way or another, here's what you have to beware of. Let's take your example, with generating set $\{x_1,x_2,x_3\}$. Ask yourself: how many partitions does this generating set have? Answer: there are 5 partitions in total: one partition into one set of cardinality 3; three partitions into two sets, one of cardinality 2 and one of cardinality 1; and one partition into three sets each of cardinality $1$.
So, if this statement was true then for any group $G$ generated by $3$ elements, and for any integer $m \ge 3$, the number of subgroups of index $m$ would be $\le 5$.
That, however, is directly contradicted by the fact that the number of subgroups of index $m$ in a rank 3 free group goes to $+\infty$ as $m \to +\infty$.
