Let $f:\mathbb{R}^{n}\rightarrow \mathbb{C}$ be a continuous function. I Believe that
$$\int_{S^{n-1}}f(x\cdot \omega)\,d\omega=\int_{S^{n-1}}f(-x\cdot \omega)\,d\omega.\qquad \qquad (1)$$
The reason is $-\omega$ is the unit vector antipodal to $\omega$, and then when we integrate on $\mathbb{S}^{n-1}$ we are summing the same integrand over the same surface.
How can we show this rigorously ?
Well, we change variables $\bar{\omega}=-\omega$, and then the Jacobian is $(-1)^n$. But we do not worry about the minus sign since we can reorient the sphere in the "correct" direction and obtain (1).
How can we write down the math?
The measure $d\omega$ is surface area measure on $S^{n-1}.$ This measure is rotation invariant. I.e., if $T$ is a rotation (aka orthogonal transformation), then
$$\int_Sf(\omega)\,d\omega = \int_S f(T(\omega))\,d\omega$$
for all such $T.$ Since $T(z)=-z$ is a rotation, we get the result.