The following theorem was stated in my algebra class today
Theorem: If a group $G$ acts on a set $\Omega$, then for each $x \in \Omega$ we have a $G$-equivariant bijection $f : G/\operatorname{Stab}(x) \to \operatorname{Orb}(x)$. In particular any transitive group action is equivalent to an action on cosets.
The last part of this theorem states that "any transitive group action is equivalent to an action on cosets", by this I assume what is mean is the following.
Suppose a group $G$ acts on $\Omega$ transitively, then by the above theorem we have $G/\operatorname{Stab}(x)$ to be a $G$-set, with the action being multiplication of cosets, $(g, y\operatorname{Stab}(x)) \mapsto gy\operatorname{Stab}(x)$ and in this case we have $\operatorname{Orb}(x) = \Omega$ to also be a $G$-set with the action being the same action as $G$ on $\Omega$, $(g, x) \mapsto gx$. We then have an equivariant bijection $f : G/\operatorname{Stab}(x) \to \Omega$, and then we can conclude that this particular action on cosets is $G$-equivariant to the group action of $G$ on $\Omega$.
Is that what is meant by the closing remark of this theorem?