Let $k$ be a field of characteristic $2$, and let $A$ be the path algebra over $k$ of the quiver with two vertices, $1$ and $2$, and arrows $a:1\to 2$ and $b:2\to 1$, 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.

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.