Is the fixed subring a symmetric algebra?

Question

Let A be a finite dimensional symmetric k-algebra over some field k. The set of units of A is denoted by U(A). Suppose G is a cyclic group of prime order which acts via inner algebra automorphism on A, say, there is a homomorphism $\phi : G \rightarrow U(A)$ such that $a\cdot g:= a ^{\phi(g)}$ for all $a\in A, g\in G$ and this action preserves scalar multiplication by k. We donote the fixed subring under the action of G by $A^G$. My question is the following:

Is $A^G$ always a symmetric k-algebra in this condition?

Answer 1

Let $k$ be a field of characteristic $2$, and let $A$ be the path algebra over $k$ of the quiver with two vertices, $v_1$ and $v_2$, and arrows $a:v_1\to v_2$ and $b:v_2\to v_1$, modulo the relations $aba=0$ and $bab=0$.

Then $A$ is a symmetric algebra (with symmetrizing form given by $\varphi(ab)=\varphi(ba)=1$ and $\varphi(p)=0$ for all paths $p$ of length $0$ or $1$).

Let $\phi(G)$ be the subgroup of $U(A)$ generated by $1+a$, which has order $2$.

Then $A^G$ is the span of $\{1,a,ab,ba\}$, which is isomorphic to $k[x,y,z]/(x,y,z)^2$ and is not symmetric.