That's a sloppy explanation.
It's more simple than that. Angular momentum and angular velocity are parallel, if your axis lies along one of the eigen-axes of the object: of course, symmetry axes are usually eigen-axes (defined by the eigenvectors of the moment of inertia tensor). That's obvious from the definition $\vec{\Gamma}=J\vec{\omega}$ which is an eigenvalue problem.
Here, we said nothing of the axis going through the center of mass, it's not required to be mounted through the center. Consider a point mass at some radius. It's obviously $\Gamma = \vec{r}\times m\vec{v}$, which is perpendicular to the plane of $\vec{r}$ and $\vec{v}$ -- the same as the axis of rotation.
Physically, two things can make your setup excentric and vibrating... if $\Gamma$ is not parallel to $\omega$, then you have oscillating TORQUE acting on the axis, which will cause vibrations and ruin your bearings. But... if the center of mass is not on the axis, then FORCES will act on the axis, as the linear momentum is not conserved (you have to spin the velocity vector of the center of mass, which requires a force). This will likewise want to ruin your bearings, just in a different way (no net torque, but plenty of forces).
In that sense... balanced rotation requires both conditions to be met, but no... center of mass position has nothing to do with the angular momentum and axis being parallel.
EDIT:
Linear algebra tells you that there are always 3 mutually perpendicular axes, each with possibly its own moment of inertia, for which the angular momentum is parallel to the angular velocity vector. This means that even for objects with no symmetry at all, these special axes exist: that's why the symmetry definition is incomplete. If you find a symmetry axis, then it is an eigen-axis, but if there is no symmetry, you must unfortunately compute the entire thing and use linear algebra to find these directions.
If inertia is equal for any pair (or all three) axes, any direction between those axes are also good. This way, a cylinder has one axis aligned with its symmetry axis, but instead of having only two other special axes (like a general blob would have), any axis, perpendicular to the long symmetry axis also has this property. Even more special case, for a sphere, cube, or a cylinder with very special aspect ratio (or cone with special aspect ratio, or a lot of other objects too, including some with no symmetry), all (not just 3 perpendicular) axes are equivalent: they all have the same moment of inertia, and they all keep the angular momentum parallel to the rotation axis.
There are a few ways to justify it.
First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly.
Another way is by direct calculus. Suppose some 1D object is made of point masses at positions $x_i$ with masses $m_i$. Then
$$I(x) = \sum m_i (x_i-x)^2$$
is the moment of inertia about $x$. The minimum is attained when the derivative is zero, so
$$0 = 2 \sum m_i (x_i - x)$$
which implies that
$$x = \frac{\sum m_i x_i}{\sum m_i}.$$
This is the definition of the center of mass.
Best Answer
The definition is incorrect as you have stated.
It is not zero, but only a minimum when axis is passing through CM.
This is because there are particles of physical mass that are away from the mathematically defined line axis through CM, as $\int r^2 \, dm = k^2 \,m. $
If we rotate about a parallel axis, we add a term $m h ^2$ making total moment of inertia $ m(k^2+h^2). $
If $h=0$ we still have the central mass.
For moment of inertia to be zero the mass of object has to be zero!
EDIT
As an example integration of an odd function area is zero, but when squared the volume of squared function is non-zero.