I have been doing algebra exercises recently and I stumbled across this problem that I struggled to solve.
Suppose a finite group $G$ acts on a finite set $A$ so that for every nontrivial $g \in G$ there exists a unique fixed point (i.e., there is exactly one $a \in A$, depending on $g$, such that $g(a) = a$). Prove that this fixed point is the same for all $g \in G$.
I had a feeling that we were supposed to use the orbit stabilizer theorem but I didn't know how. So I made the following attempt:
Suppose $|G|=n$ and suppose all $g_i\in G$ such that $g_i(a_0)=a_0\in A$ except for $m$ of them. Then $G=\{g_1,...,g_{k1},...,g_{km},...,g_n\}$ acts on $a_0$ will have the orbit $\{a_0,...,a_{k1},...,a_{km},...,a_0\}$. Then we let $G$ acts on $a_{k1}$ and get the orbit $\{a_1,a_2,...,a_0,a_{k1},...,a_n\}$. $a_0$ is in it because we have $g_{k1}^{-1}$ and $a_{k1}$ exists because these two orbits have $a_0$ in common$\Rightarrow$they are the same orbit. Moreover, $g_{k1}$ can't fix $a_{k1}$ therefore $\exists g_{kx}(a_{k1})=a_{k1}$.
This is the furthest I can get. I'm trying to force a contradiction by assuming the negative but got really stuck.
Could you give me some hints to push me forward? If this direction is totally wrong, what would be the correct path?
Thanks a lot!
The idea of GGplay in the comments works. Since all non-identity elements have exactly one fixed point, and the number $r$ of orbits is the average of the fixed points of the elements (which I believe is not a result of Burnside), we get $r=(|A|+|G|-1)/|G|$, so
$$|A| = (r-1)|G|+1.$$
Since the length of each orbit is a factor of $|G|$, there cannot be more than one orbit of length less than $|G|$, because two such orbits would have combined length at most $|G|$ making the total length at most $|G| + (r-2)|G| < |A|$.
So there must be $r-1$ orbits of length $|G|$ and one of length $1$, which proves the result.