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, i.e. the representations in the symmetric groups on $\Delta$ and $\Gamma$ are isomorphic as subgroups of the symmetric groups, and under the bijection $\psi$ corresponding elements induce the same cycles. Now 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 $$ \psi(\alpha^g) = \psi(\alpha)^g. $$ Now let $G$ act transitively on $\Omega$ and suppose $N \unlhd G$. Then some orbit $\alpha^N$ forms a block, as if $\beta^g = \gamma$ with $\beta, \gamma \in \alpha^N$ as $(\beta^n)^g = \beta^{gn'} = \gamma^n \in \alpha^N$ and $\beta^N = \alpha^N$, hence $(\alpha^N)^g \subseteq \alpha^N$. Now let $N$ be intransitive, and suppose $\Delta$ and $\Gamma$ are two orbits of $N$. As $G$ acts transitive we have $\Delta = \Gamma^g$ for some $g \in G$. Then $N$ acts on $\Delta$ and $\Gamma$ permutation isomorphic. For the isomorphism we have $\varphi : N \to N$ by $\varphi(n) = n^g$, and the bijection is $\psi(\alpha) = \alpha^g$. Then $$ \psi(\alpha^n) = \alpha^{ng} = \alpha^{gn^g} = \psi(\alpha)^{\varphi(n)}. $$ But in general the actions on the blocks are not equivalent. I am looking for examples where the actions of a normal subgroup on different blocks are indeed not equivalent? If $G$ is abelian, then the above map $\varphi : N \to N$ is the identity map, so $G$ must be non-abelian. An artificial example that comes to my mind is, if $A$ is some group with two nonequivalent transitive actions on two sets $\Delta_1, \Delta_2$, and $t$ is some involution mapping $\Delta_1$ onto $\Delta_2$, then $A \times \langle t \rangle$ acts on $\Delta = \Delta_1 \cup \Delta_2$ by the definition $\alpha^{nt} = (\alpha^n)^t$, which is well-defined as $nt = tn$ (or which could be easily seen by using the external direct product). Then $\Delta_1$ and $\Delta_2$ are two orbits of $A$ with inequivalent actions. But this example is an ad hoc construction.
So do you know any (preferable more natural) examples where a normal subgroup acts non-equivalent on its orbits?
EDIT: Derek's answer made me think about the kernels of representations. And I guess we have: If $G$ acts on $\Delta$ and $\Gamma$ and $\rho: G \to S_{\Delta}$ and $\sigma : G \to S_{\Gamma}$ are the permutation representations. Then:
i) If $G$ acts equivalent on $\Delta$ and $\Gamma$ then $\mbox{ker}(\rho) = \mbox{ker}(\sigma)$ and $G / \mbox{ker}(\rho)$ acts equivalent, or which says the same $$ \psi(\alpha^{\rho(g)}) = \psi(\alpha)^{\sigma(g)}. $$
If $G$ acts equivalent, then $$ g \in \mbox{ker}(\rho) \Leftrightarrow \forall \alpha : \alpha^g = \alpha \Leftrightarrow \forall \alpha : \psi(\alpha^g) = \psi(\alpha) \Leftrightarrow \forall \alpha : \psi(\alpha)^g = \psi(\alpha) \Leftrightarrow \forall g \in \mbox{ker}(\sigma) $$ and $\psi(\alpha^{\rho(g)}) = \psi(\alpha^g) = \psi(\alpha)^g = \psi(\alpha)^{\sigma(g)}$. If the above conditions are fulfilled, it is also easily seen that we can define two equivalent actions by $\alpha^g := \varphi^{\rho(g)}$ and $\psi(\alpha)^g := \psi(\alpha)^{\sigma(g)}$.
ii) If $G$ acts permutation isomorphic then $\mbox{ker}(\rho) \cong \mbox{ker}(\sigma)$ and the images act permutation isomorphic. Let $G$ act permutation isomorphic, a simple calculation shows $g \in \mbox{ker}(\rho) \Leftrightarrow \varphi(g) \in \mbox{ker}(\sigma)$, hence they are isomorphic by $\varphi$. Further $\varphi$ induces an isomorphism on the images by setting $\overline \varphi(\rho(g)) := \sigma(\varphi(g))$. This is well-defines, as if $\rho(g) = \rho(h)$, then \begin{align*} \psi(\alpha)^{\sigma(\varphi(g)} & = \psi(\alpha)^{\varphi(g)} \\ & = \psi(\alpha^g) \\ & = \psi(\alpha^{\rho(g)}) = \psi(\alpha^{\rho(h)}) = \psi(\alpha^h) = \psi(\alpha)^{\varphi(h)} = \psi(\alpha)^{\sigma(\varphi(h)}. \end{align*} Then \begin{align*} \varphi(\alpha^{\rho(g)}) & = \varphi(\alpha^g) \\ & = \varphi(\alpha)^{\varphi(g)} \\ & = \varphi(\alpha)^{\sigma(\varphi(g))} \\ & = \varphi(\alpha)^{\overline \varphi(\rho(g))}. \end{align*} But if it is possible to conversely built up two permutation isomorphic actions from those conditions I do not know yet, as in general $G/N \cong H/M$ with $N\cong M$ does not imply $G \cong H$.
There was a small example in one of your recent questions: $A_4$ acting transitively on $6$ points, with $|N|=4$. Then $N$ has three orbits of length $2$ with different kernels on each of the orbits.