Problem: Let $A$ be an infinite abelian group. Assume that order of every element of $A$ is power of $p$. Let G be a finite $p$ subgroup of $\operatorname{Aut}A$, and $$ A^G = \{a \in A \mid a = g(a), \forall g \in G \} . $$ Show that $\lvert A^G \rvert > 1$.
Attempt: If $A$ is finite, it is easily proven. So, if there exits finite subgoup of $A$ (call it $B$), and $g(B) = B$ for all $g \in G$, it is done ($\{1\} \subsetneq B^G \subseteq A^G$).
Question: Does such subgroup $B$ always exist? Or, anothe way to solve the problem?
Many thanks!
The first part of this answer answers the original question when $A$ was assumed finite, the second part answers the new version.
The group $G$ acts on the set $A$. There is a trivial orbit $\{Id\}$, where $Id$ is the identity element of $A$, all orbits have lengths powers of $p$ since $|G|$ is a power of $p$. The sum of all lengths is $|A|$, a power of $p$. Hence there should be another trivial orbit $\{x\}$, hence $Id\ne x\in A^G$, and $|A^G|\ge 2$.
An answer to the new version. Take $Id\ne a\in A$. Then the orbit $G\cdot a$ of $a$ under the action of $G$ is finite. Let $A_1$ be the subgroup generated by that orbit. Then $A_1$ is finite because $A$ is locally finite (being Abelian and torsion). Moreover $A_1$ is a finite $p$-group since all its elements have orders powers of $p$. The group $G$ acts on $A_1$. By the previous argument $|A_1^G|\ge 2$, Q.E.D. since $A_1^G\subset A^G$.