How can the 'spin' angular momentum vector NOT always be parallel to the spin angular velocity?

The spin angular momentum, that is the angular momentum with respect to the center of mass, is conserved if there are no torques applied on the object.

The spin angular momentum is:

$$mathbf L = int_v mathbf r times dmathbf p = int_v mathbf r times frac{dmathbf r}{dt} rho dv$$

Where $mathbf r$ is the position vector of the points of the object from the COM. Because the object is a rigid body, the distance between points is constant, including distance from the COM. So the only possible movement of all points relative to the COM is a common instantaneous rotation.

But nothing requires that this rotation must be constant in time! For each point, the infinitesimal rotation means a displacement $Delta mathbf r$

$$Delta mathbf r = Rmathbf r - mathbf r = (R - I)mathbf r implies frac{d mathbf r}{dt} = Omega mathbf r$$

Where $mathbf r$ are the position vectors relative to the selected origin in the body, $I$ is the identity matrix, $R$ is the infinitesimal rotation matrix:

begin{Bmatrix} 1 & -theta_3 & theta_2 theta_3 & 1 & -theta_1 -theta_2 & theta_1 & 1 end{Bmatrix}

and $Omega$ is the matrix:

begin{Bmatrix} 0 & -omega_3 & omega_2 omega_3 & 0 & -omega_1 -omega_2 & omega_1 & 0 end{Bmatrix}

The $omega$'s are the instantaneous angular velocities relative to the coordinates axis. The cross product in the integral of the angular momentum becomes:

$$mathbf r times frac{dmathbf r}{dt} = mathbf r times Omega mathbf r$$

Expanding the cross product, the angular momentum at any given time, relative to the point in the body, can be expressed as: $mathbf L = (int_v rho mathbf M dv) omega = mathbf Iomega$

where $mathbf I$ is the inertia matrix, $M$ is the square matrix:

begin{Bmatrix} (y^2 + z^2) & -xy & -xz –yx & (z^2 + x^2) & -yz -zx & –zy & (x^2 + y^2) end{Bmatrix}

and $omega$ is the column matrix:

begin{Bmatrix} omega_1 omega_2 omega_3 end{Bmatrix}

As can be seen, $mathbf I$ and $omega$ can change with time, keeping $mathbf L$ constant. There are however 3 axis in the object for that $mathbf L$ is indeed parallel to $omega$, corresponding to the principal moments of inertia.