Let $G$ be a group acting on $\Delta$, and $H$ be a group acting on $\Gamma$. If there exists an isomorphism $\varphi : G \to H$ and a bijection $\psi : \Delta \to \Gamma$ such that $$ \psi( \alpha^g ) = \psi(\alpha)^{\varphi(g)} $$ for all $\alpha \in \psi, g \in G$ then the two group actions are said to be permutation isomorphic. They are called equivalent iff $G = H$ and $\varphi = \mbox{id}_G$, i.e. if $G$ acts on two sets $\Delta$ and $\Gamma$ and we have a bijection $\psi : \Delta \to \Gamma$ such that $$ \psi(\alpha^g) = \psi(\alpha)^g. $$ So the conditions of equivalent actions is stronger. But why we bother about that stronger notion? Where is it essential?
This question is related to my other recent question in which I asked for examples differentiating both notions. What I see there in the example provided in the answer, that just under the notion of equivalent actions we can for example conclude more specific about individual elements, for example that if it fixes a point in one action, then also in the other. So this makes it easier to count fixed points of a single element, under permutation isomorphic actions we just know that it is isomorphic to some other element fixing a point.
But are there any other causes. Any theorems that rely specifically on the notion of equivalent action, and not just permutation isomorphic groups?
I know that two actions on the cosets of some subgroup by right multiplication are equivalent iff the two subgroups are conjugate. By using that each action of a transitive group is equivalent to an action on some subgroup, this help in classifying all equivalent actions of some group. This is a nice connection, but still does not answer why we bother about classifying them, i.e. what's their significance in the first place?
So to be more specific, do you know any theorem's were it is essential that two group actions are equivalent (either in its statement or its proof). Or any other intuitive ideas on the significance of this notion as opposed to permutation isomorphic?
Equivalence is the better condition because it comes from isomorphism in a very natural category to write down: namely, for any group $G$, there is a category $\text{Set}^G$ of $G$-sets, and two $G$-sets are isomorphic iff they are equivalent in your sense. In order to get permutation isomorphism we need to also look at the action of $\text{Aut}(G)$ on this category.
Variations of this category occur naturally in, for example, Galois theory: one way of stating the Galois correspondence is that if $k$ is a field, then the category of finite products of finite separable extensions of $k$ is equivalent to the category of finite continuous $G$-sets, where $G = \text{Gal}(k_s/k)$ is the absolute Galois group of $k$. Automorphisms of $G$ don't enter into the picture.