I have a group $G$ defined by $\langle x,y \mid x^3=y^4=1, yx=x^2y\rangle $ and a left action defined by conjucation.
I have calculated the orbits to be $G.1 = [1] $ $G.x=[x,x^2]$ $G.y=[y, xy, x^2y]$ $G.y^2=[y^2, xy^2, x^2y^2]$ and $G.y^3 = [y^3, xy^3,x^2y^3] $
However, when calculating the stabilizer subgroups of $x$ and $y$ I get $Gx=[1,x,x^2]$ and $Gy=[1,y,y^2,y^3]$
The orbit of $y$ and its stabilizer subgroup follow the orbit stabilizer theorem as multiplying their order we get $12$ which is the order of the group $G$.
But using $x$ we get $2\times 3 = 6$ instead of $12$.
What am I missing?