Is a finite group action on a finite set determined by its fixed points? Suppose I am given a finite group $G$, and a finite set $X$, and told that $G$ acts on $X$, but not told how.
However, suppose for every subgroup $H\le G$, I am given the subset $X^H\subset X$ of elements fixed by everything in $H$.
Does the above data allow us to determine the action of $G$ on $X$ up to isomorphism? Ie, could two nonisomorphic actions give rise to the same collection of fixed points $\{X^H : H\le G\}$? (Here two actions "." and "$\cdot$" of $G$ on $X$ are isomorphic if there is a self-bijection $f : X\rightarrow X$ with $f(g.x) = g\cdot f(x)$
 A: If $G$ acts transitively on $X$, and $x_{0}$ is a fixed element of $X$, your information will yield the stabilizer $G_{x_{0}}$, as the largest subgroup $H$ such that $x_{0} \in X^{H}$. So you will know that the action of $G$ on $X$ is isomorphic to that of $G$ on the cosets of $G_{x_{0}}$.
A: I believe that the answer is yes. In fact I think that it is enough to assume that the sizes of the fixed point sets of each $H \le G$ are the same in both actions.
Here is a rough argument. I think what you call isomorphic actions are usually called equivalent actions so I will do so. The aim is define an equivalence $\tau:X \to X$ between the two actions $\alpha$ and $\beta$. It is more convenient to let the two actions be on sets $X_1$ and $X_2$ with $X_1=X_2=X$.
Firstly, the fixed points of $G$ under $\alpha$  can be mapped bijectively by $\tau$ to the fixed points under $\beta$, so we can now forget about those,
and remove the fixed points under the two actions from $X_1$ and $X_2$. (So $X_1$ and $X_2$ might now be different.)
Now let $H$ be a maximal subgroup of $G$. Then, each fixed point of $H$ under an action must correspond to an orbit of the action that is equivalent to the action by multiplication on the cosets of $H$, as in Andreas Caranti's answer. There is exactly one fixed point for each such orbit. So we can now define $\tau$ on each of these orbits under $\alpha$, mapping them to corresponding orbits under $\beta$. We can do that for each of the maximal subgroups, and after doing that, we can remove the points in all of these orbits from $X_1$ and $X_2$. (In fact when we define the correspondence between two such orbits we are dealing with all maximal subgroups in a conjugacy class.)
Now we consider the two-step maximal subgroups $H$ of $G$ i.e. maximal subgroups of maximal subgroups. Since we have removed orbits in which a point stabilizer is maximal, all fixed points of such subgroups must correspond to orbits on the cosets of those subgroups. Now there could be more than one fixed point of $H$ in each orbit, but there will be the same number in each orbit, so we can define $\tau$ on these orbits as before.
Then we consider $3$-step maximal subgroups, and carry on until like that until $\tau$ is defined on the whole of $X_1$.
A: So the answer is yes, and it is equivalent (by the Galois correspondence) to the positive answer to this question:
Is a locally constant etale sheaf determined by its sections over finite etale neighbhorhoods?
