Moment of inertia tensor and symmetry of the object

Using the inertia tensor $I_{jk}$ of an object, you can construct an ellipsoid satisfying the equation $$1=x_{j}I_{jk}x_{k} .$$ As you said, the eigenvectors of $I_{jk}$ are the principal axis $vec{r}_{j}$ of your object with the respective moment of inertia $theta_{j}$ as eigenvalue. Thus, changing coordinates to the principal axis of the object diagonalises the inertia tensor and the equation for the ellipsoid is then $$1=theta_{1}x_{1}^{2}+theta_{2}x_{2}^{2}+theta_{3}x_{3}^{2} .$$ The semi-axles of this ellipsoid are parallel to the principal axles of your object and their length is the inverse square root of the moment of inertia in that direction $a_{j}=frac{1}{sqrt{theta_{j}}}$.

The inertia tensor is a bit more descriptive in the spherical tensor basis (so instead of having nine basis dyads made from combinations like $hat x hat y$, you have 9 basis tensors that transform like the nine $Y_l^m$ for $l in {0,1,2}$).

Since $I_{ij}$ is antisymmetric, all $l=1$ spherical tensors are zero. The $l=0$ portion is:

$$ I^{(0,0)} = frac 1 3 {rm Tr}(I)delta_{ij}$$

and that is the spherically symmetric part of the object.

Removing the spherically symmetric part leaves a "natural" (read: symmetric, trace-free) rank-2 tensor:

$$ S_{ij} = I_{ij} -I^{(0,0)}$$

The spherical comments are:

$$ S^{(2,0)} = sqrt{frac 3 2}S_{zz} $$

This tells you if your object is prolate or oblate.

$$ S^{(2,pm 2)} = frac 1 2 [S_{xx}-S_{yy}pm 2iS{xy}]$$

You will find that $S^{(2,+2)} = (S^{(2,-2)})^*$, and that if you are in diagonal coordinates, they are real and equal. If the value is 0, then the object is cylindrically symmetric.

$$ S^{(2,pm 1)} = frac 1 2 [S_{zx}pm iS_{zy}]$$

Here: $S^{(2,+1)} = -(S^{(2,-1)})^*$, and the term is zero in diagonal coordinates.