Let $p$ be a prime number. Let $S$ be a group of order $p^r$ and $T$ a set with $m$ elements. Let $$E=\{A\subset S\times T\ |\ |A|=p^r\}.$$ Then, the mapping $\phi:S\rightarrow\mathfrak{S}_{S\times T},\,s\mapsto((x,y)\mapsto(sx,y)),$ is a group homomorphism. Let $$\hat{\phi}:S\rightarrow\mathfrak{S}_{\mathfrak{P}(S\times T)}, s\mapsto (A\mapsto\phi_s(A)),$$ be the canonical extension of this action to $\mathfrak{P}(S\times T)$. Note that $E\subset\mathfrak{P}(S\times T)$ is stable under this operation; hence $S$ operates on $E$. Let $$E^S:=\{A\in E\ |\ (\forall s)(s\in S\implies\phi_s(A)=A)\}.$$
I want to show that $$E^S=\{Y\ |\ (\exists t)(t\in T\ \&\ Y=S\times\{t\})\}.$$
Attempt:
Let $A\in E^S$: then $A\subset S\times T$ such that $|A|=p^r$ and $\phi_s(A)=A$, for all $s\in S$. This means that $A\ne\emptyset$: let $z\in A$. I have to show that there exists $t\in T$ such that $A=S\times\{t\}$.
Is $pr_2(z)$ a good candidate for $t$? I am not able to show that $A=S\times\{pr_2(z)\}$. Alternatively, what could $t$ be?
Let $A\in E^S$ nonempty and $(x,t)\in A$. Then $x\in S$ and so for $x^{-1}\in S$ we have $$\phi_{x^{-1}}((x,t))=(e,t)\in A,$$ where $e\in S$ denotes the identity element. It follows that for every $s\in S$ we have $$\phi_s((e,t))=(s,t)\in A,$$ and so $S\times\{t\}\subset A$. Because $|A|=|S|=|S\times\{t\}|$ it follows that $A=S\times\{t\}$.
Conversely, for every $t\in T$ the set $A_t:=S\times\{t\}$ satisfies $A_t\in E^S$.