The Earth is mostly fluid. This may seem a strange claim but the rock in the mantle behaves like an extremely viscous fluid, which is why continental drift can happen.
Anyhow, if you imagine a stationary drop of liquid it will form a sphere. This is a bit of a cheat because small drops form spheres due to surface tension not gravity, but the end results are similar. If you start the drop rotating the water at the "equator" is going to feel an outwards force due to the rotation, so the drop will change shape and get bigger around the equator while the poles flatten. This shape is known as an oblate spheroid, and indeed it's the shape of the Earth because the Earth behaves like a rotating fluid drop.
To try and calculate the change of shape is a little messy, but luckily someone has done all the hard work for you and you can find the results:
Thayer Watkins: The shape of a rotating fluid mass.
The asymmetric problem gets into more complicated aspects of the kinematics of rotating bodies than are usually the point when presenting this example.
In the initial, arms out symmetric configuration, the skater's center of mass is directly over the pivot point. If she brings one arm in, and still has the axis of her body strictly vertical, then her center of mass is no longer over the pivot point, and, were she not spinning, she would fall. Now, because she is spinning, you are dealing with a problem similar to that of a gyroscope in a gravitational field whose axis of rotation is non-vertical: the gyroscope precesses.
More step wise
1. (initial condition) the skater is spinning about a vertical axis, both arms outstretched.
2. skater starts pulling her left arm inward, this changes the location of her center of mass.
3. skater starts "falling" towards her outstretched right arm.
4. this is a torque (due to gravity and the friction on the ground that keeps her skate tip at a fixed point on the ice) on a spinning body.
5. this ends up causing her main axis of rotation to precess.
I believe that I can see these kinds of effects in some of the examples here. First thing to note is that the skate is always tracing a circle on the ice. The size of this circle is related to the degree of asymmetry in the skaters body position: more asymmetric - larger circle. This is consistent with the skater needing to manage the location of her center of mass by adjusting her body, in particular her legs, and/or needing to manange rotating about a non-vertical axis in such a way that her overall angular momentum is (very close to) exactly vertical.
This page has a nice summary of rigid body mechanics, which if worked through, could be applied to this situation.
Best Answer
From conservation of angular momentum we have $(I+\Delta I)(\omega+\Delta \omega) = I\omega,$ or $$\frac{\Delta \omega}{\omega} = - \frac{\Delta I}{I+\Delta I} \simeq -\frac{\Delta I}{I}.$$ We make the following simplifying assumptions:
The earth is a sphere of uniform density of mass $M$ and radius $R$.
The building is constructed on the equator by digging out a sphere of earth of mass $m$ and radius $r$ and raising it a distance $2r$. We assume $r\ll R$.
With these assumptions we find $$\begin{eqnarray*} \frac{\Delta \omega}{\omega} &\simeq& - \frac{m(R+r)^2-m (R-r)^2}{\frac{2}{5}M R^2} \\ &\simeq& - 10\frac{r^4}{R^4}. \end{eqnarray*}$$
Assuming that $r$ is 200 m (the geometric mean of 100 m, 100 m, and 750 m) we find
$$\begin{eqnarray*} \frac{\Delta \omega}{\omega} &\simeq& -10^{-17}. \end{eqnarray*}$$ Atomic clocks are accurate to about one part in $10^{14}$, so there is no hope in measuring such an effect.