Relation of Angular and Linear velocity with radius of circular path

Linear/tangential velocity, in a circlular path, increases with the increase in radius and decreases with the decrease in radius. Hence, the angular velocity remains the same no matter what the change in radius is(W=V/r). However, when we talk about the conservation of angular momentum, we say that since the momentum is conserved, as we increase the radius, the linear velocity must decrease to keep it constant(because L=mvr), which concludes it is inversely related to the radius, which concludes that angular velocity is inversely propotional to the square of radius, ( when an ice skater brings his arms inwards, it rotates with greater angular velocity)but It is clear from what's stated above, that angular velocity must remain the same, regardless of what the radius is, isn't it?

$$\vec{l} = mrv\vec{e}_z = mr^2\omega \vec{e}_z,$$ which I have written here directly depending on the angular velocity $\omega$. So you see that a point particle has a fixed angular velocity $\omega$ as long as $l$, $r$ and $m$ are fixed. To change it, a torque has to be applied. Actually, it is defined as the change of angular momentum ($D = \dot{l}$).
When you are referring to the ice skater, it is important that this deals with the rotation of a rigid body, i.e. a system of individual point masses. The total angular momentum of this system is obtained by integrating (or for a discrete set of particles, summed) over all infinitesimal contributions of the point particles. This obviously depends on the geometry of the object. The information about the object is contained in the so-called tensor of inertia, $$\vec{L} = I \cdot \vec{\omega}$$ To put a long story short, the ice-skater increases $\omega$ when contracting herself because on the system as a whole, there is no torque. Let's assume the ice-skater is a cylinder, for which one can find $I = \frac{m}{2}r^2$, so that $L = \frac{m}{2} r^2 \omega$. When the ice-skater is reducing her radius, we find a $\omega$-gain of $$\frac{\omega_2}{\omega_1} = \frac{r_1^2}{r_2^2}.$$