Let $S$ be a surface in $R^3$, for which coordinate vector field of $S$ has zero mean on $S$. Assume that for any vector $n$, a normal plane to $n$ exist, such that $S$ is symmetric about it.
How can we see that for any direction $n$, the plane through the origin normal to $n$ is a plane of symmetry?
Thought: Denote the plane through the origin perpendicular to $N_p$ by $P_p$. We need to show that $P_p$ is a plane of symmetry.
For any point $p\in S$, choose $\Pi_p$, a plane of symmetry perpendicular to $p$. Then the reflection $R(p)$ of $p$ across $\Pi_p$ is also in $S$. Moreove, the tangent plane to $R(p)$ is $R(T_pS)$. Note here that the argument $p$ indicates which reflection is applied, the plane of reflection changes with $p$.
If $\Pi_p=\{x\in\mathbb R^3:\langle x,N_p\rangle=\eta(p)\}$, where $N$ is a global Gauss map, it would suffice to show that we can choose $\Pi_p:\eta(p) = 0$. Alternatively, choose $\Pi_p:(0,0,0)\in\Pi_p$.
I would instead look at $\Pi_n$, the planes of symmetry perpendicular to the vector $n$. I don't think you need tangent planes or Gauss maps.
For each vector $n$, find $\Pi_n$, and look at $\int_{S^+} p\,dS$ and $\int_{S^-} p\,dS$, where $S^+$ is the half of $S$ above $\Pi_n$ and $S^-$ the half below. Can you conclude that the origin must lie on $\Pi_n$?
EDIT:
Call $c^+ = \int_{S^+} p\,dS$, etc. Then as you say $c^- = R(c^+)$, where $Rq = q - 2nn^T(q-b)$ is reflection about $\Pi_n$ and $b$ is a point on $\Pi_n$. Then you know from the problem statement that $c^-+c^+ = 0$, so $$2c^+ - 2nn^T(c^+-b) = 0.$$ Take the dot product of both sides by $n$ to get $$n\cdot b = 0$$ and $\Pi_n$ passes through the origin.