Consider finite groups $H\subset G$. Denote by $H\backslash G$ the right orbit space. That is: $$H\backslash G = \{H\sigma_1,\dots,H\sigma_n\}$$ for some choice of $\sigma_1,\dots,\sigma_n \in G$. Consider the right action of $G$ on $H\backslash G$. If we fix a $g\in G$, this will decompose $H\backslash G$ into a number of equivalence classes like so: $$\{H\sigma_1,\dots,H\sigma_1g^{f_1(g)}\},\dots,\{H\sigma_n,\dots,H\sigma_ng^{f_n(g)}\}.$$ Assume that $H,G$ are such that for all $g\in G$, $f_i(g)$ is constant in $i$. Does this imply that $H$ is normal in $G$?
Note that the converse is true. If $H$ were normal in $G$, then right cosets correspond to left cosets and therefore, we can identify the above right action of $G$ with the left action of $G$ in which case $f_i(g)$ is simply the order of $g$ modulo $H$.
Yes it's equivalent. If the subgroup is not normal then choose $g \in H \setminus f^{-1}Hf$ for some $f \in G$. Then the orbit of $H$ under $g$ has length $1$, but the orbit of $Hf$ has length greater than $1$. There is no need to assume that the groups are finite.i