This is to answer a natural and interesting question raised by Joseph O'Rourke's in a comment above. Indeed we have:

Any open subset $A$ of $\mathbb{R}^n$, star-shaped wrto a point $P$ and which is partitioned into two pieces of equal measure by each hyperplane through $P$, is center-symmetric wrto $P$.

Here below I'm describing (with some freedom) the main argument of the proof, that I extracted by this short paper by K.J.Falconer, for those who have no access to Jstor. The key-point is a consequence of the Funk-Hecke theorem: the integral mean of a spherical harmonic $S$ over the hemi-sphere centered at $\theta\in \mathbb{S}^m$, as a function of $\theta$, is a non-zero scalar multiple of $S$ (the F-H theorem says much more; so I think hopefully there is also a short proof of this fact).

Assume $P$ is the origin, and let $f:\mathbb{S}^{n-1}\to \mathbb{R}_+$ describe the boundary of $A$ in polar coordinates (that is, for $x=r\theta \in \mathbb{R}^n$ with $r\ge 0$ and $\theta\in \mathbb{S}^{n-1}:=\partial B _ {\mathbb{R}^n }(0,1 )$, then $x\in A$ if and only if $r< f(\theta)\\ $ ).

By integrating in polar coordinates, the condition on $A$ writes:

$$\int_{(\psi\cdot \theta)\ge0} f(\theta)^n d\theta=\int_{(\psi\cdot \theta)\ge0} f(-\theta)^n d\theta\\ , \quad\forall\psi\in\mathbb{S}^{n-1} $$ and we are to show that this implies that $f^n$, thus $f$ itself, is an even function (the other implication is of course quite obvious, and reflects the fact that a center-symmetric $A$ is equi-partitioned by any hyperplane through the origin). To this end, consider the transformation $u\in L^2(\mathbb{S}^{n-1})\mapsto \tilde u\in L^2(\mathbb{S}^{n-1})$ defined by

$$\tilde u (\psi):=\int_{(\psi\cdot \theta)\ge0} u(\theta)d\theta=\int_\mathbb{{S}^{n-1}} \chi_ { \mathbb{R}_+}(\psi\cdot \theta) u(\theta) d\theta\\ , \quad\forall\psi\in\mathbb{S}^{n-1} \\ . $$

Due to the symmetry of the integral kernel $\chi_ { \mathbb{R}_+}(\psi\cdot \theta)$ we have $(\tilde u\cdot v) _ {L^2}=(u\cdot \tilde v) _ {L^2}$; moreover, as recalled, spherical harmonics are eigenfunctions of this transformation, with non-zero eigenvalues. Therefore, if $\tilde u$ is even then for any odd spherical harmonic $S$ we have $(u\cdot \tilde S) _ {L^2}=(\tilde u\cdot S) _ {L^2}=0$, thus also $(u\cdot S) _ {L^2}=0$, and $u$ is an even function too, ending the proof.

Moreover, just differentiating with respect to $\psi$ it is easy to see that for a star-shaped open subset $A$ the above property of equipartition by all hyperplanes is equivalent to: any section of $A$ by a hyperplane through $P$ has $n-1$ dimensional barycenter located in $P$ (note that in dimension 2 this immediately implies that $A$ is center-symmetric).

This is to answer a natural and interesting question raised by Joseph O'Rourke's in a comment above. Indeed we have:

Any open subset $A$ of $\mathbb{R}^n$, star-shaped wrto a point $P$ and which is partitioned into two pieces of equal measure by each hyperplane through $P$, is center-symmetric wrto $P$.

Here below I'm describing (with some freedom) the main argument of the proof, that I extracted by this short paper by K.J.Falconer, for those who have no access to Jstor. The key-point is a consequence of the Funk-Hecke theorem: the integral mean of a spherical harmonic $S$ over the hemi-sphere centered at $\theta\in \mathbb{S}^m$, as a function of $\theta$, is a non-zero scalar multiple of $S$ (the F-H theorem says much more; so I think hopefully there is also a short proof of this fact).

Assume $P$ is the origin, and let $f:\mathbb{S}^{n-1}\to \mathbb{R}_+$ describe the boundary of $A$ in polar coordinates (that is, for $x=r\theta \in \mathbb{R}^n$ with $r\ge 0$ and $\theta\in \mathbb{S}^{n-1}:=\partial B _ {\mathbb{R}^n }(0,1 )$, then $x\in A$ if and only if $r< f(\theta)\, $ ).

By integrating in polar coordinates, the condition on $A$ writes:

$$\int_{(\psi\cdot \theta)\ge0} f(\theta)^n d\theta=\int_{(\psi\cdot \theta)\ge0} f(-\theta)^n d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1} $$ and we are to show that this implies that $f^n$, thus $f$ itself, is an even function (the other implication is of course quite obvious, and reflects the fact that a center-symmetric $A$ is equi-partitioned by any hyperplane through the origin). To this end, consider the transformation $u\in L^2(\mathbb{S}^{n-1})\mapsto \tilde u\in L^2(\mathbb{S}^{n-1})$ defined by

$$\tilde u (\psi):=\int_{(\psi\cdot \theta)\ge0} u(\theta)d\theta=\int_\mathbb{{S}^{n-1}} \chi_ { \mathbb{R}_+}(\psi\cdot \theta) u(\theta) d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1} \, . $$

Due to the symmetry of the integral kernel $\chi_ { \mathbb{R}_+}(\psi\cdot \theta)$ we have $(\tilde u\cdot v) _ {L^2}=(u\cdot \tilde v) _ {L^2}$; moreover, as recalled, spherical harmonics are eigenfunctions of this transformation, with non-zero eigenvalues. Therefore, if $\tilde u$ is even then for any odd spherical harmonic $S$ we have $(u\cdot \tilde S) _ {L^2}=(\tilde u\cdot S) _ {L^2}=0$, thus also $(u\cdot S) _ {L^2}=0$, and $u$ is an even function too, ending the proof.

Moreover, just differentiating with respect to $\psi$ it is easy to see that for a star-shaped open subset $A$ the above property of equipartition by all hyperplanes is equivalent to: any section of $A$ by a hyperplane through $P$ has $n-1$ dimensional barycenter located in $P$ (note that in dimension 2 this immediately implies that $A$ is center-symmetric).