Find an example of sets of cosets of different cardinality

$G:=S_5$ and $H:={(1),(23),(24),(34),(234),(243)}cong S_3$. Set $x:=(35)$ and $y:=(13)(45)$. We have begin{align} &Hx={(35),(253),(24)(35),(345),(2534),(2453)},\ &Hy={(13)(45),(132)(45),(13)(254),(1354),(13542),(13254)}. end{align} The set of cosets ${Hxhmid hin H}$ has $3$ elements and they are begin{align} &{(35),(253),(24)(35),(345),(2534),(2453)},\ &{(235),(25),(2435),(2345),(25)(34),(245)},\ &{(345),(2543),(2354),(45),(254),(23)(45)}; end{align} the set of cosets ${Hyhmid hin H}$ has $6$ elements and they are begin{align} & {(13)(45),(132)(45),(13)(254),(1354),(13542),(13254)},\ &{(123)(45),(12)(45),(12543),(12354),(12)(354),(1254)},\ &{(13)(245),(13452),(13)(25),(13524),(1352),(134)(25)},\ &{(1453),(14532),(14253),(14)(35),(142)(35),(14)(253)},\ &{(14523),(1452),(143)(25),(14)(235),(14352),(14)(25)},\ &{(12453),(12)(345),(1253),(124)(35),(12)(35),(12534)}. end{align}

Hence we are done.

Consider the symmetry group $D_8$ of a square. Let $C_2$ denote the subgroup fixing a corner $x$ (so $C_2$ consists of the identity, and reflection through the diagonal containing $x$). Then we can identify the $4$ corners of the square with the cosets of $C_2$. That is all the elements of $C_2g$ map $x$ to $xg$, so we may identify the coset $C_2g$ with the corner $xg$, for each $gin D_8$.

The orbits of corners under $C_2$ have different sizes: one orbit is the two corners adjacent to $x$, another is the single corner opposite to $x$.

Thus if $a$ is a $90^circ$ rotation, then ${C_2ah|hin C_2}$ is two cosets, whilst ${C_2a^2h|hin C_2}$ is one.

First consider $G=S_3$ and $H={e,(1 2)}$, with $x=e$ and $y=(1 2 3)$. Then $$ {Hxhmid hin H} = {H} quadtext{while}quad {Hyhmid hin H} = {H(1 2 3), H(1 2)}. $$

But wait, you say, we're not allowed to take $xin H$? This isn't actually that serious a restriction, since for any nontrivial group $K$ and any $kin Ksetminus{e}$, we can now replace $G$ by $Gtimes K$ and $H$ by $Htimes{e}$, and $x$ and $y$ by $xtimes k$ and $ytimes k$.