$H_1,H_2$ are groups of order $128$, $X_1,X_2$ are sets of order $8$ and $H_i$ acts faithfully on $H_i$ for $i=1,2$. Show that there is an isomorphism of groups $\pi: H_1\to H_2$ and a bijection of sets $\phi:X_1\to X_2$ so that $\phi(g\cdot x)=\pi(g)\cdot\phi(x)$ for all $g\in H_1$ and $x\in X_1$.
By Sylow's theorem it's easy to see $H_1,H_2$ are isomorphic. But it seems hard for me to express the solution of second part clearly.
Here is my attempt: Since $H_1, H_2$ can be embedded into $S_8$ and are isomorphic, we can identify them as a same subgroup of $S_8$, say $H$. Then $H_i$ acts on $X_i$ is the same as $H$ acts on $X_i$, which gives the index of elements of $X_i$ from $1$ to $8$, thus induces a bijection $\phi: X_1\to X_2$, s.t. $\phi(h\cdot x)=h\cdot\phi(x)$, for any $h\in H$. Therefore by the identification of $h, g, \pi(g)$, with $g\in H_1$, we have $\phi(g\cdot x)=\pi(g)\cdot\phi(x)$.
Another attempt is: Since $S_8$ has an $8$ cycle, $H_1$ must contain an $8$-cycle, say $g_1$. Define $g_2=\pi(g_1)$, thus $g_2$ is also a $8$-cycle. Assume $X_1=\{a_1,...,a_8\}$, with $g_1\cdot a_i\to a_{i+1}$, $g_1\cdot a_8\to a_{1}$. And $X_2=\{b_1,...,b_8\}$, with $g_2\cdot b_i\to b_{i+1}$, $g_2\cdot b_8\to b_{1}$.
Define $\phi:X_1\to X_2$, $a_i\mapsto b_i$. This implies that $\pi(g_1^n\cdot a_i)=g_2^n\cdot b_i=g_2^n\cdot\pi(a_i)$.
Now assume $x=a_i$, $g\cdot a_i=a_j$, thus $\phi(g\cdot a_i)=b_j$. And $\phi(x)=\phi(a_i)=b_i$.
$g\cdot a_i=g_1^k\cdot a_i$, thus $\pi(g\cdot a_i)=\pi(g_1^k\cdot a_i)=g_2^k\cdot\pi(a_i)=\phi(g_1^k)\cdot\pi(a_i)$.
But I can't go further.
The actions of $H_i$ on $X_i$ correspond to homomorphisms $\alpha_i: H_i\to {\rm Sym}(X_i)$, and the fact that the actions are faithful says that each $\alpha_i$ is injective.
Let $\tau:X_1 \to X_2$ be an arbitrary bijection. This induces a natural isomorphism $\bar{\tau}: {\rm Sym}(X_1) \to {\rm Sym}(X_2)$, such that $\bar{\tau}(g)(\tau(x)) = \tau(g(x))$ for all $x \in X_1$ and $g \in G_1$.
Now $\bar{\tau}(\alpha_1(H_1))$ and $\alpha_2(H_2)$ both have order $2^7$, so they are Sylow $2$-subgroups of ${\rm Sym}(X_2)$. Hence they are conjugate by an element $t \in {\rm Sym}(X_2)$; i.e. $t\bar{\tau}(\alpha_1(H_1))t^{-1} = \alpha_2(H_2)$.
Now we can define the isomorphism $\pi:H_1 \to H_2$ by $\pi(h_1) = h_2$, where $t\bar{\tau}(\alpha_1(h_1))t^{-1} = \alpha_2(h_2)$.
We define the bijection $\phi:X_1 \to X_2$ by $\phi(x) = t\tau(x)$.
The condition we are trying to prove is then $t\tau(\alpha_1(h_1)(x)) = \alpha_2(\pi(h_1))(t\tau(x))$ for all $x \in X_1$ and $h_1 \in H_1$.
The right hand side of the required equality is $$t\bar{\tau}(\alpha_1(h_1))t^{-1}(t\tau(x))= t\bar{\tau}(\alpha_1(h_1))(\tau(x))= t\tau(\alpha_1(h_1)(x)),$$ as required.
The idea of the proof is not so hard - it seemed somehow clear that $t\tau$ should be the equivalence $\phi$ - but it took me far too long to get the details right!