Suppose we have the following cones: \begin{align} C_1&=\{(x_1,x_2,x_3): x_1 \le x_2 \le x_3 \} \\ C_2&=\{(x_1,x_2,x_3): x_1 \le x_3 \le x_2 \} \\ C_3&=\{(x_1,x_2,x_3): x_2 \le x_3 \le x_1 \} \\ C_4&=\{(x_1,x_2,x_3): x_2 \le x_1 \le x_3 \} \\ C_5&=\{(x_1,x_2,x_3): x_3 \le x_1 \le x_2 \} \\ C_6&=\{(x_1,x_2,x_3): x_3 \le x_2 \le x_1 \} \end{align} Note that the collection $C_1,\ldots, C_6$ is a partition of $\mathbb{R}^3$.
Now, let $A \in \mathbb{R}^{3 \times 3}$ be symmetric and positive definite matrix. Suppose that the following sequence of equation holds \begin{align} &{\rm vol} \left(C_1 \cap A B(0,1) \right)={\rm vol} \left(C_2 \cap A B(0,1) \right)={\rm vol} \left(C_3 \cap A B(0,1) \right)\\ &={\rm vol} \left(C_4 \cap A B(0,1) \right)={\rm vol} \left(C_5 \cap A B(0,1) \right)={\rm vol} \left(C_6 \cap A B(0,1) \right) \end{align} where $B(0,1) \subseteq \mathbb{R}^3$ is a ball centered at $0$ of radius $1$, and ${\rm vol}()$ is a volume operator.
Note that $A B(0,1)$ is an ellipse. Therefore, in words, given a partition of the space by symmetric cones, in how many ways can we orient an ellipse such that intersection with each cone has the same volume.
Question: What are all possible matrices $A$ (which are symmetric and positive definite) that satisfy the above condition?
For $=3$, I think there are two types of solutions: 1) is a diagonal matrix, 2) all $2$ eigenvalues of $A$ are the same and one can be arbitrary. In words, the solution is a ball or, an ellipse stretched all the direction where all cones meet.
The boundaries of the $6$ regions $C_k$ lie on $3$ planes: $x_1 = x_2$, $x_2 = x_3$ and $x_3 = x_1$. These $3$ planes intersect at a common line $\ell : x_1 = x_2 = x_3$. If we look at these planes from direction $\hat{w} = \frac{1}{\sqrt{3}}(1,1,1)$, they become $3$ lines intersecting at origin with angle $60^\circ$ among them.
Let $P$ be the plane $x_1 + x_2 + x_3 = 0$. It has $\hat{w}$ as a normal vector. Given any ellipsoid $\mathcal{E}$ centered at origin, its orthogonal projection to $P$ is an ellipse $E$. If we apply a Steiner symmetrization to $\mathcal{E}$ along direction $\hat{w}$, we obtain another ellipsoid $\mathcal{E}'$ centered at origin with $\mathcal{E}' \cap P = E$. Since $\hat{w}$ is tangent to above $3$ planes, we have
$${\rm vol}(C_k \cap \mathcal{E}) = {\rm vol}(C_k \cap \mathcal{E}')\quad\text{ for all } k.$$ Let $2\beta$ be the length of the line segment $\mathcal{E} \cap \ell = \mathcal{E'} \cap \ell$. It is not hard to show $${\rm vol}(C_k \cap \mathcal{E}') = \frac{4\beta}{3}{\rm area}(C_k \cap E)$$
This means in order for all $C_k \cap \mathcal{E}$ to have equal volume, we need all $C_k \cap E$ to have equal area. An obvious sufficient condition for this to happen is $E$ is a circle. It turns out this condition is also necessary.
When $E$ is not a circle, choose a coordinate system $(x,y)$ on plane $P$ to make $E$ has the standard form: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \quad\text{ with }\quad \lambda \stackrel{def}{=} \frac{a}{b} \ne 1 $$ For any $\theta \in [0,\frac{\pi}{3})$, let $t_j = \tan(\theta + \frac{j\pi}{3})$ for $j = 0,1,2$. By construction, they satisfy $$\frac{k_1 - k_0}{1 + k_1 k_0} = \frac{k_2 - k_1}{1 + k_2 k_1} = \frac{k_0 - k_2}{1 + k_0 k_2} = \tan\frac{\pi}{3} = \sqrt{3}$$ If the three lines $y = k_0 x, y = k_1 x, y = k_2 x$ split $E$ into six pieces of equal area, then the three lines $y = \lambda k_0 x, y = \lambda k_1 x, y = \lambda k_2 x$ split a circle with radius $a$ into six pieces of equal area. The angle between these 3 new lines is $60^\circ$ again. This leads to $$\frac{\lambda(k_1 - k_0)}{1 + \lambda^2k_1 k_0} = \frac{\lambda(k_2 - k_1)}{1 + \lambda^2 k_2 k_1} = \frac{\lambda(k_0 - k_2)}{1 + \lambda^2 k_0 k_2} = \sqrt{3}$$ Notice for any $(i,j) = (1,0), (2,1), (0,2)$, the equality $$\frac{k_i - k_j}{1 + k_i k_j} = \sqrt{3} = \frac{\lambda(k_i - k_j)}{1 + \lambda^2k_i k_j}$$ can be simplified to $\lambda k_i k_j = 1$ when $\lambda \ne 1$. This forces $k_0 = k_1 = k_2$ which is impossible. This means when $E$ is not a circle, we cannot find $3$ lines intersecting at $60^\circ$ to split $E$ into six pieces of equal area.
Extend $\hat{w}$ to an orthonormal basis: $$\hat{u} = \frac1{\sqrt{6}}(2,-1,-1), \hat{v} = \frac1{\sqrt{2}}(0,1,-1), \hat{w} = \frac1{\sqrt{3}}(1,1,1)$$ For any point $\vec{x} = (x_1,x_2,x_3)$, let $(u,v,w)$ be its coordinates with respect to this basis.
$$\vec{x} = u\hat{u} + v\hat{v} + w\hat{w}\quad\iff\quad \begin{cases} u &= \frac{1}{\sqrt{6}}(2x_1 - x_2 - x_3)\\ v &= \frac{1}{\sqrt{2}}(x_2 - x_3)\\ w &= \frac{1}{\sqrt{3}}(x_1 + x_2 + x_3) \end{cases}$$
In order for all $C_k \cap \mathcal{E}$ to have equal area, above arguments tell us $E$ is a circle and in coordinate system $(u,v,w)$, the equation of $\mathcal{E}'$ takes following form:
$$\frac{u^2 + v^2}{\alpha^2} + \frac{(w - \gamma u - \delta v)^2}{\beta^2} \le 1$$ where $\alpha$ is the radius of $E$ and $\gamma, \delta$ controls the amount of shearing one need to bring $\mathcal{E}'$ to $\mathcal{E}$.
As one can see, there is no need for $A$ to have any double eigenvalue. I will leave the translation of this equation of $\mathcal{E}$ to coefficients of $A$ for you.