The **total** charge density, which is the sum of free charge density and bound charge density is constant all over the sphere, but the free and bound charge densities differ for upper and lower half of the sphere.

Your question about symmetry of total charge density can be answered easily, assuming that you know the symmetry of electric field. (your reasoning for it's spherical symmetry is correct):

($E_1$ means electric field in the first region and $E_2$ the field of the second region)

$$\mathbf{E}_1=\mathbf{E}_2 =\mathbf{E}_{out} \tag{as you stated}$$
$$\mathbf{E}_{out} . \hat r-\mathbf{E}_{in}.\hat r=\frac{\sigma}{\epsilon}_0 \,\,\,,\,\,\,\mathbf{E}_{in}=0 \to \sigma={\epsilon}_0 \mathbf{E}_{out} . \hat r$$

You can arrive at this result by calculating free and bound charge densities directly by solving for potential using Laplace equation too:

The problem is clearly azimuthal symmetric, and also according to your argument, it isn't a function of $\theta$ neither; i.e., the problem is 1-D and $V$ is a function of $r$.

Using the fact that potential is continuous at the boundary of the two regions and also vanishes at $\infty$, the only remaining term of the potential expansion in spherical coordinates will be:

$$V=\frac{Cq}{r}$$
$$ \mathbf{D}=-\epsilon \nabla V \to \cases{\mathbf{D}_1=\frac{\epsilon_1Cq}{r^2}\hat r \\ \mathbf{D}_2=\frac{\epsilon_2Cq}{r^2}\hat r}$$

Now, assuming a spherical surface surrounding the sphere and applying Gauss law for $\mathbf{D}$, we will find $C$:

$$\oint{\mathbf{D}.d\mathbf{S}}=Q_{free}\to C=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)}$$

$$V=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r}$$

Now we find $\mathbf{D}_1$ , $\mathbf{D}_2$ and $\mathbf{E}$ :

$$\mathbf{E}=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

$$\mathbf{D}_1=\frac{\epsilon_1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

$$\mathbf{D}_2=\frac{\epsilon_2}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

using the relation $\sigma_f=\mathbf{D}.\hat r $ we will have:

$$\to \cases{\sigma_{1f}=\frac{q\epsilon_1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{1}{R^2} \\ \sigma_{2f}=\frac{q\epsilon_2}{2\pi (\epsilon_1 + \epsilon_2)} \frac{1}{R^2}}$$

From these two relations for $\sigma_{f}$s it is evident that the total free charge is equal to $q$, as expected:
$$\sigma_{1f}\times 2\pi R^2+\sigma_{2f}\times 2\pi R^2=q$$

Now we find bound charge densities. First we should find polarization vectors $\mathbf{P}_1$ and $\mathbf{P}_2$:

$$\mathbf{P}=\mathbf{D}-\epsilon_0 \mathbf{E} \to \cases{\mathbf{P}_1= \frac{\epsilon_1- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r \\ \mathbf{P}_2=\frac{\epsilon_2- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r }$$
and

$$\sigma_b =\mathbf{P}.\hat r\to \cases{\sigma_{b1}=-\frac{\epsilon_1- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{R^2} \\ \sigma_{b2}=-\frac{\epsilon_2- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{R^2} }$$

Now you can see that on each half sphere the sum of bound and free charge densities is equal, as expected:
$$\sigma_{b1}+\sigma_{f1}=\sigma_{b2}+\sigma_{f2}$$

The answer by @BrianMoths is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation.

Start with your charge distribution and a "guess" for the direction of the electric field.

As you can see, I made the guess have a component upward. We'll see shortly why this leads to a contradiction.

Now do a "symmetry operation," which is a fancy phrase for "do something that leaves something else unchanged. In this case, I'm going to reflect everything about a horizontal line. I mean *everything.*

The "top" of the sheet became the "bottom." This is just arbitrary labeling so you can tell I flipped the charge distribution. The electric field is flipped too. (Imagine looking at everything in a mirror, and you'll realize why things are flipped the way they are.)

Hopefully, everything is okay so far. But now compare the original situation with the new inverted one.

You have exactly the same charge distribution. You can't tell that I flipped it, except for my arbitrary labeling. But if you have the same charge distribution, you ought to also have the same electric field. As you can see, this is not the case, which means I made a mistake somewhere.

The only direction for the electric field that does not lead to this contradiction is perpendicular to the sheet of charge.

## Best Answer

It's helpful that the posted image is sideways, because it already shows the answer: as you rotate the coordinate system, and advance by the same distance, $r$, everything looks exactly the same. This is true if you rotate in the plane of the table, or if you hold the page against a wall and rotate it.

If instead of rotating the coordinate system you were to imagine holding a smooth, uniformly colored sphere, like a ball bearing, you will find that no matter how you rotate it, it will appear the same.

This is spherical symmetry. Since everything is the same, the intensity calculated at one point will be the same as that for any other point. However, the radial vectors will all point in different directions - they all point directly away from the center, each one towards its own point on the surface.