Group action on Artin ring

Question

Let $k$ be a field and $A$ is a finite dimensional commutative $k$-algebra. Suppose a finite group $G$ acts on $A$ by automorphisms. Then $G$ acts on $\operatorname{Spec}(A)$. Suppose $A^G=\{a \in A | ga=g \text{ for any } g \in G \}$ is isomorphic to $k$. Is action on $\operatorname{Spec}(A)$ transitive under such assumptions?

Answer 1

Yes, this is true. Partition $\operatorname{Spec} A$ in to orbits $O_1,\cdots,O_n$ under the $G$ action. Then on each orbit $O_i$, define the regular function $f_i$ which assigns to each point in the orbit the value $1$ and $0$ to all other points. Then each $f_i$ is invariant under the $G$ action on $A$ and all $f_i$ are mutually orthogonal idempotents, so $A^G\cong \bigoplus_{i=1}^{i=n} Af_i$, and each $Af_i$ contains a copy of $k$. If $A^G\cong k$, then $n=1$ so there's only one orbit and thus $G$ acts transitively on $\operatorname{Spec} A$.