A volume form on the sphere which gives equal areas to all hemispheres is invariant under the antipodal map? Let $\omega$ be a volume form on $\mathbb{S}^2$ with the property that the induced area (w.r.t $\omega$) of all the hemispheres is the same.
Is it true that $\omega$ is invariant under the antipodal map? i.e let $f(x)=-x$, does $f^*\omega=\omega$ hold?
The assumption implies that for any hemisphere $A \subseteq \mathbb{S}^2$, we have
$$ \int_{A}\omega=\int_{f(A)}\omega=\int_{A}f^*\omega=\frac{1}{2}\int_{\mathbb{S}^2} \omega. \tag{1}$$
Edit:

Lemma 1:
$\omega$ satisfies $(1)$ if and only if $\int_{A}L_X\omega=0$ for every  Killing field $X$ and hemisphere $A$.

Proof:
$\Rightarrow$ Suppose $\omega$ satisfies $(1)$:
Let $\phi_t$ be the flow of a Killing field $X$ on $\mathbb{S}^2$, and let $A$ be a hemisphere. Since all the $\phi_t(A)$ are hemispheres we get
$$ \int_{\phi_t(A)}\omega=\int_{A}\phi_t^*\omega=\text{const},$$ so
$$ 0=\frac{d}{dt}|_{t=0} \int_{A}\phi_t^*\omega=\int_{A}\frac{d}{dt}|_{t=0}\phi_t^*\omega=\int_{A}L_X\omega.$$
$\Leftarrow$ Suppose $\omega$ satisfies $\int_{A}L_X\omega=0$ for every  Killing field $X$ and hemisphere $A$:
Let $A$ be a hemisphere. We want to show $ \int_{A}\omega=\int_{-A}\omega$. There exist a Killing field $X$ s.t its flow takes $A$ to $-A$ at some time $t=t_0$. (i.e if $\phi_t$ is the flow, $\phi_{t_0}(A)=-A$). Now,
$$ \frac{d}{dt}|_{t=s} \int_{A}\phi_t^*\omega=\int_{A}\frac{d}{dt}|_{t=s}\phi_t^*\omega=\int_{A}\phi_s^*L_X\omega=\int_{\phi_s(A)}L_X\omega=0,$$ where the last equality is exactly the assumption.
Since $s$ was arbitrary, this implies $\int_{A}\phi_t^*\omega=\int_{\phi_t(A)}\omega$ is independent of $t$, so in particular $$\int_{A}\omega=\int_{\phi_0(A)}\omega=\int_{\phi_{t_0}(A)}\omega=\int_{-A}\omega,$$
as required.


Lemma 2:
The conditions in lemma 1 are equivalent to
$\int_{C} i_X\omega=0$
for any great circle $C$ and Killing field $X$.

Proof:
Let $C$ be a great circle. $C$ bounds a hemisphere $A$.
By Cartan's magic formula, $L_x\omega=d(i_X\omega)$, so
$$ \int_{A}L_X\omega=\int_{A} d(i_X\omega)=\int_{\partial A} i_X\omega=\int_{C} i_X\omega.$$
Since any hemisphere has a great circle for a boundary we are done.

Discussion:
Let $\tilde \omega$ be the standard round volume. Let $\omega=h\tilde \omega$ be an arbitrary form. $\omega$ is invariant iff $h(x)=h(-x)$.
By looking at a great circle $C$, we see that the condition in lemma 2 is equivalent to
$$ \int_C   h|_{C}(\theta)\sin \theta=0,  \int_C   h|_{C}(\theta)\cos \theta=0.$$
This is because the space of Killing fields on $\mathbb{S}^2$ is $3$-dimensional, and one non-trivial Killing field always fixes the two hemispheres bounded by $C$, so we are effectively left with two equations.
Now, as mentioned by  Anthony Carapetis,  there is a solution which is not $\pi$-periodic: We can just take $h$ to be constant along latitude's, and for latitude $\theta$, set $h(\theta)=2+\sin(3\theta)$.
However, this solution cannot be lifted to a suitable candidate on the sphere.
 A: Any signed measure $\mu$ on the sphere which gives equal areas to all hemispheres is invariant under $x\mapsto -x.$
We can assume:


*

*$\mu$ is odd, by taking the odd part $\mu'(B)=\frac{\mu(B)-\mu(-B)}2.$

*(Optional, if $\mu$ is already assumed to be equal to a continuous function) $\mu$ is equal to a polynomial in $\mathbb R^3,$ i.e. a linear combination of spherical harmonics, by convolution by a polynomial on $SO(3)$ (using Riesz representation + Weierstrass approximation to ensure there is some polynomial that gives a non-zero convolution)

*$\mu$ depends only on the latitude, by rotating to ensure the north pole is non-zero, and averaging over rotations fixing the north pole


The value on each $y$ co-ordinate give an odd continuous function $p:[-1,1]\to \mathbb R.$ We need to show that $p$ is identically zero.
By considering a great circle whose $y$-co-ordinate ranges from $-a$ to $a$, if I have calculated correctly, the Killing field argument implies the condition
$$\int_0^{2\pi} p(a\cos t) \cos t dt=0\qquad\text{ for all }0\leq a\leq 1.$$
The function on the sphere defined by $q(y)=p(y)y$ is even, continuous, and has identically zero Funk transform, so it is itself identically zero. This implies $\mu$ is zero. (In fact this argument doesn't even need any averaging. With the averaging, it is also possible to use the Abel transform - see the comments.)
Alternatively, since we can assume $p$ is analytic, suppose $p(x)=Cx^k+O(x^{k+2})$ for some odd $k.$ Then $\int_0^{2\pi}p(a\cos t)\cos t dt=Ca^{k}\int_0^{2\pi}\cos^{k+1}(t)dt + O(a^{k+2})$ as $a\to 0,$ so $C=0.$ This proves that $p$ is identically zero, hence so is the measure $\mu.$
And without the odd assumption, we get that $\mu$ must be even.
