Angular Momentum of a rigid, extended object: When we see a rotating object, is the state of rotation is totally relative? Angular momentum of an object is a physical quantity that depends on the chosen point about which to calculate the angular momentum.
It is often said that an object that has been thrown up in the air and is rotating, is physically rotating about the center of mass. I don't think that is what is physically happening. We chose the center of mass as the point of rotation because it is convenient mathematically (it makes the separation of translation and rotational energy easier).
We could choose any other point (inside, outside the object, in motion or at rest relative to the object) and calculate the rotation about that arbitrary point.
It is not a physical fact that the free object rotates about the center of mass.
Even an object that is constrained to rotate about a fixed axis, we could describe the rotation about any point, not necessarily points on the fixed, constrained axis.
 A: 
So when we see an object rotating, its state of rotation is totally relative, as it happens for many other physical quantities...
  Is that correct?

The state of rotation is not $totally$ relative; for example,  the angular velocity of rotation is the same for all points of reference. It is true that you can use different point of reference for the rotation from the center of mass, and sometimes this is useful, especially if that point moves with constant velocity. But often it is the center of mass that moves with constant velocity, hence it is said

It is often said that an object that has been thrown up in the air and is rotating, is physically rotating about the center of mass.

because in inertial reference frame, only the center of mass can be viewed as non-rotating.
A: *

*Angular momentum is an intrinsic quantity at the center of mass $\vec{H}_C = \bar{I}_C \vec\omega$, where $\bar{I}_C$ is the 3×3 inertia tensor at the center of mass C.


*Angular momentum measured at any other point must include the effect of linear momentum $\vec{L}$ as well. $\vec{H}_A = \vec{H}_C + \vec{r}_{AC} \times \vec{L} = \bar{I}_C \vec\omega + \vec{r}_{AC} \times m \vec{v}_C$.


*For example if we fix the rotation about point A with $\vec{v}_C = \vec{\omega}\times \vec{r}_{AC}$ then the angular momentum about A is $\vec{H}_A = \bar{I}_C \vec\omega + \vec{r}_{AC} \times m \left(\vec{\omega}\times \vec{r}_{AC} \right)$ which is known as the parallel axis theorem when written as
$$ \vec{H}_A = \bar{I}_A \vec{\omega} = \left( \bar{I}_C - m \vec{r}_{AC} \times \vec{r}_{AC} \times \right) \vec{\omega}$$
See my answer to a related question for more details.


*The free state of a rigid body is a rotation + parallel translation of the center of mass (also known as screw motion), but even a loaded rigid body will still undergo screw motion, just about a different axis. The choice of the center of mass, is not one of convenience, but a natural one that comes out of the equations of motion. In fact, there is a geometric relationship between the axis of a force and the resulting axis of rotation discovered by Sir Robert Stawell Ball called Screw Theory


*Here are some statements about rigid body motion that might explain why the center of mass is "special":

*

*A pure translation is achieved by a force through the center of mass

*A pure torque will rotate a body about its center of mass (dual statement to above)

*A pure translation yields no angular momentum

*A body with only angular momentum will rotate about its center of mass (dual statement to the above)



*All your questions can be answered from the following fundamental relationships:

*

*The motion screw of rigid body (at the center of mass) is defined by $$v_C = \begin{pmatrix} \vec{\omega} \\ \vec{v}_C \end{pmatrix} $$ where $\vec{v}_C$ is the linear velocity of the c.m. and $\vec{\omega}$ the angular velocity of the body.


*The momentum screw of a rigid body is defined by $$ \begin{aligned} L_C & = I_C v_C  \\
 \begin{pmatrix} \vec{L} \\ \vec{H}_C \end{pmatrix} & = \begin{bmatrix} 0 & m \\ \bar{I}_C & 0 \end{bmatrix} \begin{pmatrix} \vec{\omega} \\ \vec{v}_C \end{pmatrix}= \begin{pmatrix} m \vec{v}_C \\ \bar{I}_C \vec{\omega} \end{pmatrix} \end{aligned}$$


*The sum of the force/moment screws equal to the time derivative of momentum screw $$ \begin{aligned} \sum f & = \frac{{\rm d}}{{\rm d}t} L_C \\
\sum \begin{pmatrix} \vec{F} \\ \vec{\tau}_C \end{pmatrix} & = \frac{{\rm d}}{{\rm d}t}  \begin{pmatrix} m \vec{v}_C \\ \bar{I}_C \vec{\omega} \end{pmatrix}  = \begin{pmatrix} m \vec{a}_C \\ \bar{I}_C \vec{\alpha} + \vec{\omega}\times \bar{I}_C \vec{\omega} \end{pmatrix} \end{aligned} $$




*The motion screw has the following geometrical properties:

*

*Direction: $\vec{e} = \frac{ \vec{\omega}}{|\vec{\omega}|}$

*Center of rotation (w/ respect to C): $\vec{r} = \frac{\vec{\omega}\times\vec{v}_C}{\vec{\omega} \cdot \vec{\omega}}$

*Pitch (parallel translation per rotation) $ h = \frac{ \vec{\omega} \cdot \vec{v}_C}{\vec{\omega} \cdot \vec{\omega}}$



*The force screw has the following geometrical properties:

*

*Direction: $\vec{e} = \frac{ \vec{F}}{|\vec{F}|}$

*Force axis position (w/ respect to C): $\vec{r} = \frac{\vec{F}\times\vec{\tau}_C}{\vec{F} \cdot \vec{F}}$

*Pitch (parallel torque per force) $ h = \frac{ \vec{F} \cdot \vec{\tau}_C}{\vec{F} \cdot \vec{F}}$



*Similarly for the momentum screw. The location of the momentum axis $\vec{r} = \frac{\vec{L}\times\vec{H_C}}{\vec{L} \cdot \vec{L}}$ is where a pure impulse is going to result in a pure rotation about A. This is also known as the percussion axis.


*Finally the transformation laws of screws, can move the quantities from the center of mass C to another point A such that $$ \begin{aligned} 
v_A & = X_{AC} v_C \\ 
\begin{pmatrix} \vec{\omega} \\ \vec{v}_A \end{pmatrix} & = \begin{pmatrix} \vec{\omega} \\ \vec{v}_C + \vec{r}_{AC}\times\vec{\omega} \end{pmatrix} \\
L_A & = X_{AC} L_C \\ 
\begin{pmatrix} \vec{L} \\ \vec{H}_A \end{pmatrix} & = \begin{pmatrix} \vec{L} \\ \vec{H}_C + \vec{r}_{AC}\times\vec{L} \end{pmatrix} \\
f_A & = X_{AC} f_C \\
\begin{pmatrix} \vec{F} \\ \vec{\tau}_A \end{pmatrix} & = \begin{pmatrix} \vec{F} \\ \vec{\tau}_C + \vec{r}_{AC}\times\vec{F} \end{pmatrix} 
\end{aligned} $$ with $$ X_{AC} = \begin{bmatrix} \bf{1} & \bf{0} \\ \vec{r}_{AC}\times & \bf{1} \end{bmatrix} $$ the 6×6 screw transformation matrix.
When you notice all the similarities between the motion, force, and momentum screws you can start to understand the geometry behind rigid body motion.
