Counter example about convolution of symmetric functions on locally compact groups This question is motivated on what I think is an error in a result (Lemma 1.6.5) given in Dietmar, A. and Echterhoff, S., Principles of Harmonic Analysis, 2nd edition, Springer, pp.24

Suppose $G$ is a locally compact topological group and let $\lambda$ be a (left) Haar measure. One statement of the  aforementioned Lemma is that if $\phi$ and $\psi$ are $L_1(G,\lambda)$ symmetric functions ($\phi(x^{-1})=\phi(x)$ and $\psi(x^{-1})=\psi(x)$ for all $x\in G$), then $\phi*\psi$ is also symmetric.

I think this is not necessarily true, unless $G$ is also Abelian, or $\phi=\psi$. A simple deduction one can make about $\phi*\psi$ is that
\begin{align}
\phi*\psi(x)&:=\int_G \phi(y)\psi(y^{-1}x)\,\lambda(dy)\\
&=\int_G\phi(xy)\psi(y^{-1})\,\lambda(dy)\\
&=\int_G\phi(y^{-1})\psi(yx)\Delta(y^{-1})\,\lambda(dy)\\
&=\int_G\phi(xy^{-1})\psi(y)\Delta(y^{-1})\,\lambda(dy)
\end{align}
where $\Delta$ is the modular function of $G$. From this, and the symmetry of $\phi$ and $\psi$ one can establish that
$$\phi*\psi(x^{-1})=\int_G\phi(x^{-1}y)\psi(y^{-1})\,\lambda(dy)=\int_G\phi(y^{-1}x)\psi(y)\,\lambda(dy)=\psi*\phi(x)$$
I seems to me that this is the best one can say about the between relation $\phi*\psi(x)$ and $\phi*\psi(x^{-1})$ under the general assumptions above.
Of course, if $G$ were in addition commutative, then $\phi*\psi=\psi*\phi$ and thus, $\phi*\psi$ would be symmetric.
The problem is to prove or disprove (via a counter example) whether it is indeed the case that  $\phi*\psi$ is symmetric under the general assumptions on $G$.
 A: Let $\mathbb{F}_2$ be the free group with two generators $a$ and $b.$ The functions $$\varphi=\delta_a+\delta_{a^{-1}},\quad \psi=\delta_b+\delta_{b^{-1}}$$ are symmetric.
Observe that for any $x,y\in \mathbb{F}_2$ (or in any discrete group $G$) we have $\delta_x*\delta_y=\delta_{xy}.$ Thus
$$\varphi*\psi=\delta_{ab}+\delta_{ab^{-1}}+\delta_{a^{-1}b}+\delta_{a^{-1}b^{-1}}$$
The resulting function is not symmetric.
Another example: let $G=\mathbb{Z}_2*\mathbb{Z}_2$ be the free product of two  groups $\mathbb{Z}_2.$ The group $G$  has two generators $a$ and $b$, with $a^2=e$ and $b^2=e$ as the only relations. Then $\delta_a$ and $\delta_b$ are symmetric but $\delta_a*\delta_b=\delta_{ab}$ is not symmetric as $ab\neq ba=(ab)^{-1}.$
A: I found a counterexample that show that  $\phi*\psi$ may fail to be symmetric even when $\phi$ and $\psi$ are. I am posting a solution  to share with the community.
Consider the symmetric group $S_3$ (composition as product) equipped with the discrete topology. This is a non Abelian finite topological group,  and the counting measure on the power set $\mathcal{P}(S_3)$ is a Haar measure for $S_3$. The group $S_3$ is generated by the cycles $\sigma_1=(12)$, $\sigma_2=(13)$.  Take the functions $\phi=\mathbf{1}_{\{\sigma_1\}}$ and $\psi=\mathbf{1}_{\{\sigma_2\}}$.  Since $\sigma^{-1}_j=\sigma_j$ for all $j$, both $\phi$ and $\psi$ are symmetric. For any $\sigma\in S_3$
\begin{align}
\phi*\psi(\sigma)&= \sum_{\sigma\in S_3}\mathbf{1}_{\{\sigma_1\}}(\tau)\mathbf{1}_{\{\sigma_2\}}(\tau^{-1}\sigma)=\mathbf{1}_{\{\sigma_1\sigma_2\}}(\sigma)\\
\phi*\psi(\sigma^{-1})&=\mathbf{1}_{\{\sigma_1\sigma_2\}}(\sigma^{-1})=\mathbf{1}_{\{\sigma_2\sigma_1\}}(\sigma)
\end{align}
Since $\sigma_1\sigma_2=(132)\neq\sigma_2\sigma_1=(123)$, $\phi*\psi(\sigma)\neq\phi*\psi(\sigma^{-1})$.

A counter example for a non Abelian infinite group is still welcome.
