Fixed points of an action of a group $G$ on a set $A$, where $|G|=pq$ and $|A|=p^2-2$, with $p$ and $q$ twin primes. Let $p$, $q$ be two twin primes, that is such that $q=p+2$.
Let $G$ be a group of order $pq$. Let $A$ be set of order $p^2-2$.
Suppose there is an action
$G \times A \to A$
$(g,a) \mapsto g \cdot a $
Prove that $Fix(G)$ is non-empty, where $Fix(G)=\{ a\in A| g \cdot a=a \,,\forall g \in G \}$
Since there are primes involved, I tried something with Sylow theorems:
Let $n_p(G)$ and $n_q(G)$ be, respectively, the number of $p$-Sylow and $q$-Sylow of $G$.
We must have, because $|G|=pq$, that
$n_p(G) \equiv 1\bmod q\;$,$\;n_q(G) \equiv 1\bmod p$
and
$n_p(G)|q\;$,$\;n_q(G)|p$.
This means that $n_p(G)=n_q(G)=1$. Let $Syl_p(G)=\{P\}$ and $Syl_q(G)=\{Q\}$ the set of $p$-Sylow and $q$-Sylow of G. Let's consider $P$. Since all of $p$-Sylow are conjugated, it means that $\forall g \in G$, we must have $P^g=P$, because $P$ is the only $p$-Sylow in $G$. It follows that $P$ is a normal subgroup of $G$. The same applies to $Q$.
Let's now consider the set $PQ=\{xy|x \in P, y \in Q\} \subseteq G$.
Since both $P$ and $Q$ are normal in $G$, we have that $PQ \leqslant G$.
Because $|P|=p$ and $|Q|=q$, with $p$, $q$ primes, we have that $P \cap Q = \{1\}$.
We know that $|PQ|= \frac{|P||Q|}{|P \cap Q|}$, since everything is finite. By substituting what we know, we get
$\;$$|PQ| = \frac{|P||Q|}{|\{1\}|} = \frac{pq}{1} = pq = |G|$
Thus $PQ \leqslant G$ and $|PQ|=|G|$, it must be then $PQ=G$.
Putting together these facts, we can conclude that $G \simeq P \times Q$.
Now, we know that $|P|=p$ and $|Q|=q$, which means that $P \simeq C_p$ and $Q \simeq C_q$, where $C_p$ and $C_q$ are the cyclic groups of order $p$ and $q$, respectively. But $(p,q)=1$, because of primality, it follows then that $C_p \times C_q \simeq C_{pq}$. We can write then $G \simeq C_{pq}$. Since $C_{pq}$ is cyclic, it must be abelian, which implies that $G$ must be abelian.
This is what I could do for now.
I think I should use this formula in some way, but I don't really know how
$$|A|=|Fix(G)|+\sum_{\delta \in \Delta} |G:Stab_G(\delta)|$$
Where $\Delta$ is the set containing the representatives of the non-trivial orbits of $G$, for its action on $A$, $Stab_G(\delta)=\{g \in G|g \cdot \delta = \delta \}$ and $|G:Stab_G(\delta)|$ is the index of $Stab_G(\delta)$ in $G$ which, thanks to the Orbit-Stabilizer Theorem, is equal to $|\mathcal O(\delta)|$, the order of the orbit of $\delta$, $\forall \delta \in \Delta$.
 A: You are almost there.
$$|A|=|\text{Fix}(G)|+\sum_{\delta \in \Delta} |G:\text{Stab}_G(\delta)|$$
Here, suppose for a contradiction that $|\text{Fix(G)}| = 0$. Then, we have
$$|A| = p^2-2 =\sum_{\delta \in \Delta} |G:\text{Stab}_G(\delta)| $$
But as you suggested, by Orbit Stabilizer Theorem, $|\mathcal O(\delta)|$ must divide $|G| = p(p+2)$. So, we have $|\mathcal O(\delta)| =p,p+2\text{ or }p(p+2), \forall \delta \in \Delta$. Notice that $|\mathcal O(\delta_0)| = p(p+2)$ for some $\delta_0 \in \Delta$ reaches to a contradiction since $p(p+2) > p^2-2 = |A|$. So, there are only two possibilities left for $|\mathcal O(\delta)|$, which are $p$ and $p+2$.
Now, suppose there are $a$ many orbits of size $p$, $b$ many orbits of size $p+2$. Then, we must have
$$ap+b(p+2) = p^2-2 \implies ap+bp+2b+2 = p^2 \implies p \mid (2b+2)$$
Here, since $p$ and $q$ are twin primes, $p \ne 2$. So, by Euclidean Lemma,
$$p \mid 2(b+1) \implies p \mid (b+1)$$
Now, $b < p$ since otherwise, we would have $b(p+2) \ge p^2+2p > p^2-2 = |A|$. Thus, only possibility is $b = p-1$. Then we have
$$ap +b(p+2) = ap + p^2+p-2 =p^2-2 \implies p(a+1) = 0$$
This is a contradiction.
