Relation between rotation vector derivative and angular velocity when the rotation angle is constant

$$\def\va{\vec{\alpha}} \def\vw{\vec{\omega}} \def\vn{\vec{n}}$$Let $$\va(t)$$ be a rotation vector such that its direction is the rotational axis and its length $$\alpha=|\va|$$ is the angle describing the rotation. In Is there a formula for the rotation vector in terms of the angular velocity vector? the formula $$\vec{\omega}= \dot{\vec{\alpha}} + \frac{1 - \cos \alpha}{\alpha^2} \left(\vec{\alpha} \times \dot{\vec{\alpha}}\right) + \frac{\alpha - \sin \alpha}{\alpha^3} \left(\vec{\alpha} \times \left(\vec{\alpha} \times \dot{\vec{\alpha}}\right)\right)\,$$ is given which relates angular velocity $$\vw$$ to the rotation vector $$\va$$ and its time derivative.

If I multiply the formula with $$\va$$, the two elaborate terms on the right disappear, because both contain a cross product of $$\va$$ such that their dot product with $$\va$$ is zero. I get:

$$\vw\cdot\va = \dot\va\cdot\va \tag{1}$$

Because $$\vw$$ and $$\va$$ are parallel, we also have $$\omega \alpha = \dot\va\cdot\va \tag{1}$$

Now let $$\va(t) = \alpha(t)\cdot \vn(t)$$ for unit vector $$\vn(t)$$. Then we get $$\dot\va(t) = \dot\alpha(t) \cdot\vn(t) + \alpha(t)\cdot\dot\vn(t)\,.$$

In the case $$\alpha(t)=const$$, and leaving out the $$(t)$$ for better readability, this simplifies to $$\dot\va = \alpha\cdot\dot\vn$$ and inserting into (1) results in

\begin{align} \omega\alpha = \vw\cdot\va &= \alpha \va \cdot\dot\vn\\ &= \alpha^2 \vn\cdot\dot\vn \end{align} Dividing by $$\alpha$$ we get

$$\omega = \alpha\vn\cdot\dot\vn\,.$$

Since $$\vn(t)$$ is a unit vector for every $$t$$, any change of $$\vn(t)$$ must always only change its direction, never its length, which means that $$\vn\cdot\dot\vn=0$$ and therefore

$$\omega = 0$$

in the case $$\dot\alpha=0$$, even for $$\dot\vn\neq0$$.

Question: How can this be that the axis of rotation changes its direction but $$\omega$$ and thereby angular momentum is zero? One of my assumptions of how $$\va$$ works is probably wrong. Yet all I have assumed is that $$\va$$ is just three numbers that change over time and that it can be decomposed into $$\alpha\vn$$. OK and that this is in line with the formula cited for $$\omega$$. Where is the mistake? Or can I have a change in rotation axis without having angular momentum?

Early on you make the assumption that $$\vec{\omega}$$ and $$\vec{\alpha}$$ are parallel. This is not in general true.
This may have arisen from a basic misconception about the meaning of $$\vec{\alpha}(t)$$. The direction of $$\vec{\alpha}(t)$$ is not the instantaneous axis of rotation. The axis-angle variables give you the rotation necessary to obtain the current orientation of a body (for example, at time $$t$$) relative to a reference orientation (for example, at $$t=0$$). It is true that any orientation can be expressed in this way: a rotation $$\alpha(t)$$ about a unit vector $$\vec{n}(t)$$, which we can put together as a combined vector $$\vec{\alpha}(t)\equiv \alpha(t)\vec{n}(t)$$. But this rotation depends on the entire history of the trajectory up to time $$t$$, and as was made clear in the page you referenced, Is there a formula for the rotation vector in terms of the angular velocity vector?, the relation is quite complicated. There is no particular reason why the axis $$\vec{n}(t)$$ describing the current orientation should have any relation to the direction of the current angular velocity $$\vec{\omega}(t)$$.
One can conceive of a special case where this is true: it is the simple one where $$\vec{n}$$ has been constant throughout the trajectory, and both $$\vec{\omega}$$ and $$\vec{\alpha}$$ have been parallel to $$\vec{n}$$ for the whole time. So $$\vec{\omega}(t)=\omega(t)\vec{n} \qquad\text{and}\qquad \vec{\alpha}(t)=\alpha(t)\vec{n}$$ Then the angle of rotation $$\alpha(t)$$ is just the time integral of the magnitude of the angular velocity $$\omega(t)$$. In this case, though, things are less interesting. The second and third terms of your first equation vanish identically, so $$\vec{\omega}=\dot{\vec{\alpha}}\qquad\text{and}\qquad \omega=\dot{\alpha}$$ Your derivation is correct until we get to the special case $$\alpha(t)=$$ constant, which of course correctly implies $$\omega(t)=0$$. But that is as one would expect, for this special case, it doesn't illustrate anything awry.