Some properties of the generalised Petersen Graph I was studying symmetries when I encountered this problem

Consider the graph $G$ whose vertices are labelled by subsets of order $2$ of $\{1,2,3\dots n\}$ where $n\geq5$. Two vertices are connected by an edge if and only if the subsets are disjoint. Any permutation of $\{1,2,3\dots n\}$ gives a symmetry of $G$.
a. Let $p_i=(i,i+1)$. Show that a symmetry of $G$ is determined by the images $p_i^\prime$ of $p_i$ (hint: first characterize $(j,j+2)$ in terms of the $p_i$'s. For this, consider the graph whose vertices are those not connected to either $p_j$ or $p_{j+1}$ and whose edges come from $G$).
b. Show that there are at most $n!$ symmetries.
c. Where does this proof fail for $n=4$?

I don't have much work to show for this because I couldn't make much progress. I can understand that $G$ is a generalisation of the Petersen graph (i.e., the case $n=5$) which has $5!=120$ symmetries. And I know that this generalised graph has a beautiful property that every symmetry of $G$ is induced by a permutation. For part (b), I counted that there are $\binom n 2$ choices for $p_1^\prime$, $2n-4$ choices for $p_2^\prime$, $n-3$ choices for $p_3^\prime$, etc. Also, for part (c), I know that it is possible to find four $2$-element subsets $\mathbb S_1$, $\mathbb S_2$, $\mathbb S_3$, $\mathbb S_4$ of $\{1,2,\dots ,n\}$ such that $\mathbb S_i \cap \mathbb S_j$ is a singleton set for all $i,j\in \{1,2,3,4\}$, while this property doesn't hold for any number greater than four. Is this property of any help? I can't make much progress. Please help me to proceed.
Note: as pointed out in the comments, the notation $p_i=(i,i+1)$ may create some confusions. Please note that it represents a point and NOT a symmetry.
 A: For a vertex $v \in V(G)$, let $N(v) \subseteq V(G)$ denote the set of neighbours of $v$, and $N[v] = N(v) \cup \{v\}$.
The hint is trying to get you to do the following. Consider the graph $G \setminus ( N[p_j] \cup N[p_{j+1}] )$ obtained from $G$ by removing $p_j$ and $p_{j+1}$ and all their neighbours. This amounts to removing $p_j$ and $p_{j+1}$ and all vertices corresponding to $2$-element sets $S$ such that $S \cap \{j,j+1\} = \varnothing$ or $S \cap \{j+1,j+2\} = \varnothing$. So the remaining vertices correspond to $2$-element sets $S$ such that $|S \cap \{j,j+1\}| = 1$ and $|S \cap \{j+1,j+2\}| = 1$. There are two kinds of sets of this type:

*

*sets of the form $\{j+1,k\}$ with $k \notin \{j,j+2\}$;

*the set $\{j,j+2\}$.

Let $W \subseteq V(G)$ be the collection of all vertices corresponding to sets of the first type, and let $q_j \in V(G)$ be the special vertex corresponding to the (unique) set of the second type. Then:
$$ V(G) \setminus (V[p_j] \cup V[p_{j+1}]) = W \cup \{q_j\}. $$
Any two sets $\{j+1,k\}$ and $\{j+1,k'\}$ of the first type have non-empty intersection:
$$ \{j+1,k\} \cap \{j+1,k'\} = \{j+1\}. $$
Therefore these two are not joined by an edge in $G$. In other words, there are no edges between vertices in $W$.
On the other hand, the special set $\{j,j+2\}$ of the second type is disjoint from all sets of the first type, so the special vertex $q_j$ is adjacent to all elements of $W$. Therefore $G \setminus (N[p_j] \cup N[p_{j+1}])$ is a star graph with central vertex $q_j$.
Now let $f$ be a symmetry of $G$, and let $p_i' := f(p_i)$ for all $i$. Since $f$ is a symmetry, it preserves all graph-theoretic properties. In particular, the image of the subgraph $G \setminus (N[p_j] \cup N[p_{j+1}])$ under $f$ is the subgraph $G \setminus (N[p_j'] \cup N[p_{j+1}'])$. Therefore $G \setminus (N[p_j'] \cup N[p_{j+1}'])$ is a star graph with central vertex $f(q_j)$. This shows that $f(q_j)$ can be recovered uniquely from $p_j'$ and $p_{j+1}'$, via the following algorithm:

*

*remove $p_j'$ and $p_{j+1}'$ and all their neighbours from $G$;

*the remaining graph is a star;

*let $f(q_j)$ be the central vertex of this star.

This is what the hint was trying to tell you. I'll leave the rest of the proof up to you. (Hint: it's similar, but slightly different.)
