Why does a distant rotating body have angular momentum about the origin? In a frame where the center of mass is stationary, the angular momentum of a body is invariant to translation of the origin. That is: $\mathbf{L} = \sum_i \mathbf{r_i} \times \mathbf{p_i}$ is constant regardless of choice of origin so long as $\sum_i \mathbf{p}_i = 0$.
On the one hand, this is analogous to the invariance of linear momentum under a translation of the origin.
But on the other hand, my intuition is that angular momentum is something like "momentum about a point." It is surprising that (provided the center of mass is stationary) it does not matter where the point is located.
For instance, consider the angular momentum of a distant body measured with respect to a point located "here" at the origin. If the distant body's CoM is stationary with respect to the origin, then the angular momentum is the same as if we located the origin at the CoM.
This is despite the fact that if the origin is located at the CoM, the body truly is rotating about the origin, whereas if the body is spinning at a distance, in a sense it is "stationary" with respect to the origin.
This feels surprising. Is there a useful way to think of distant "internal" rotation of a body as similar to rotation of the body about an external origin?
 A: Your observations are correct, and they rely on the premise that $\boldsymbol{p}=0$. This means that for all observers (summation point) the body is purely rotating without any translation.
The conclusion drawn is the for a purely rotating body the angular momentum measured is the same regardless of where the summation point is.
Now angular momentum is also  the moment of momentum of a point particle, such as $\boldsymbol{L} = \boldsymbol{r}_C \times \boldsymbol{p}$, as well as for a non rotating, translating rigid body.
For the general case of a rotating and translating rigid body, angular momentum is the combination of the moment of momentum term and the intrinsic angular momentum $\boldsymbol{L}_C = \mathrm{I}_C\, \boldsymbol{\omega}$ which is what you call the "internal" angular momentum.
$$ \boldsymbol{L} = \mathrm{I}_C\, \boldsymbol{\omega} + \boldsymbol{r}_C \times \boldsymbol{p} \tag{1}$$
Here $\boldsymbol{r}_C$ is the location of the center of mass relative to the summation point
Obviously, there are cases like you pointed out where $\boldsymbol{L} = \mathrm{I}_C \,\boldsymbol{\omega}$ regardless of the summation point. If the body is not translating, if the center of mass is on the summation point and if the body is translating radially outwards from the summation point.
The more interesting question here is what does (1) tell us if we were to treat the rotating-translating rigid body as an equivalent particle, moving at some distance $\boldsymbol{r}_X$ with some momentum $\boldsymbol{p}$.
This exists in the form of a line that goes through a point $\boldsymbol{r}_X$ and is parallel to momentum $\boldsymbol{p}$. I will show that angular momentum resolved anywhere along this line is strictly parallel to momentum with a proportionality factor $h$ (called the pitch) such that $\boldsymbol{L}_X = h \,\boldsymbol{p}$.
This line in space is often called the axis of percussion.
What I am going to prove here is that there always is a point $\boldsymbol{r}_X$ and pitch $h$ which would allow us to decompose the general angular momentum vector $\boldsymbol{L}$ as follows
$$ \boldsymbol{L} = h\,\boldsymbol{p} + \boldsymbol{r}_X \times \boldsymbol{p} \tag{2}$$
To find these quantities take the dot and cross product of the two momentum vectors
$$\require{cancel} \left.\begin{aligned}\boldsymbol{p}\cdot\boldsymbol{L} & =h\,\left(\boldsymbol{p}\cdot\boldsymbol{p}\right)\\
\boldsymbol{p}\times\boldsymbol{L} & =\boldsymbol{p}\times\left(\boldsymbol{r}_{X}\times\boldsymbol{p}\right)\\
 & =\boldsymbol{r}_{X}\left(\boldsymbol{p}\cdot\boldsymbol{p}\right)-\boldsymbol{p}\left(\cancel{\boldsymbol{r}_{X}\cdot\boldsymbol{p}}\right)
\end{aligned}
\right\} \begin{aligned}h & =\frac{\boldsymbol{p}\cdot\boldsymbol{L}}{\|\boldsymbol{p}\|^{2}}\\
\boldsymbol{r}_{X} & =\frac{\boldsymbol{p}\times\boldsymbol{L}}{\|\boldsymbol{p}\|^{2}}
\end{aligned} \tag{3}$$
Here $\boldsymbol{r}_X \cdot \boldsymbol{p} = 0$ because there is at least one point along the axis of percussion that is closest to the origin making $\boldsymbol{r}_X$ perpendicular to the line.
You can plug those back into (2) and see how the equations check out.
So any moving rigid body has a line in space whose a particle with the same momentum $\boldsymbol{p}$ exhibits an equivalent (or maybe equipollent is a better term) angular momentum.
There is a beautiful relationship between the location of the center of mass and the location of the axis of percussion. Without loss of generality place the origin on the rotation axis, with z-axis along the rotation and x-axis towards the center of mass

If the body has a radius of gyration $R_{\rm gyr}$ for the rotation axis $\boldsymbol{\omega}$ about the center of mass then the distance to the percussion axis $\| \boldsymbol{\ell} \|$ is described by the following equation
$$ \| \boldsymbol{c} \| \| \boldsymbol{\ell} \| = R_{\rm gyr}^2 $$
Note that $\boldsymbol{\ell}$ is always on the opposite side of the center of mass. Also, there are two special cases

*

*$\boldsymbol{c}=0$, body rotating about the center of mass, means that $\boldsymbol{\ell} = \infty$

*$\boldsymbol{\ell} =0$, body purely translating, means that $\boldsymbol{c} = \infty$ or the rotation center is at infinity.

In the scenario above you have
$$\begin{aligned}\boldsymbol{r}_{C} & =\boldsymbol{\hat{x}}c\\
\boldsymbol{\omega} & =\boldsymbol{\hat{z}}\,\omega\\
\boldsymbol{\omega}\times\boldsymbol{r}_{C} & =\omega \,c\,\boldsymbol{\hat{y}}
\end{aligned}$$
and
$$\begin{aligned}\boldsymbol{p} & =m\left(\boldsymbol{\omega}\times\boldsymbol{r}_{C}\right)=m\omega c\left(\boldsymbol{\hat{z}}\times\boldsymbol{\hat{x}}\right)=m\omega c\boldsymbol{\hat{y}}\\
\boldsymbol{L} & =\mathrm{I}_{C}\boldsymbol{\omega}+\boldsymbol{r}_{C}\times\boldsymbol{p}=\omega\left(\mathrm{I}_{C}\boldsymbol{\hat{z}}+mc^{2}\boldsymbol{\hat{z}}\right)
\end{aligned}$$
Consider the case where $\boldsymbol{\hat{z}}$ is one of the principal axes of rotation, with radius of gyration $R_{\rm gyr}^2$, then from above
$$ \boldsymbol{L}=\omega\,m\left(R_{{\rm gyr}}^{2}+c^{2}\right)\boldsymbol{\hat{z}} $$
and if you use this into (3) you will get
$$ \boxed{ \boldsymbol{r}_{X} = \boldsymbol{r}_{C}+\frac{R_{{\rm gyr}}^{2}}{c}\boldsymbol{\hat{x}}} \tag{4} $$
The axis of percussion is beyond the center of mass by a distance of $R_{\rm gyr}^2/c$.
A point particle has zero radius of gyration, so the percussion axis passes through the center of mass. The higher the radius of gyration is the further away from the center of mass the percussion axis is.
Also given a radius of gyration, the closer the center of mass is to pivot point, the further away the percussion axis is, and the other way around. The mimimum distance the axis of percussion is from the pivot point is $2 R_{\rm gyr}$ which occurs when the center of mass is one radius of gyration away from the pivot $c = R_{\rm gyr}$.
A: Consider an inertial system and the angular momentum of a system of particles about the origin of that system. The angular momentum of the system of particles is equal to the sum of two terms: (1) the angular momentum of the center of mass (CM) plus (2) the angular momentum of motion about the CM. For example, see the text Classical Mechanics by Goldstein.
If the center of mass is at rest with respect to the origin, then- as you say- the angular momentum of the system of particles is independent of the origin.  That is, the first term is zero so the angular momentum is only that of the particles relative to the CM which is independent of the origin.
Suggest you also look at Validity of rotational Newton's second law in a changing instantaneously inertial frame on this site.
A: I can only see this fact as a consequence of the definition of the angular momentum as a cross product, that has the distributive property.
For a system of $N$ particles, the angular momentum with respect to the COM is the sum: $\mathbf L = \sum_i \mathbf r_i \times \mathbf p_i$. The angular momentum with respect to another point: $\mathbf L' = \sum_i \mathbf R_i \times \mathbf p_i'$.
$\mathbf R_i = \mathbf R_{COM} + \mathbf r_i$, and if there is no relative movement between the external point and the COM, $\mathbf p_i = \mathbf p_i'$
$$\mathbf L' = \sum_i (\mathbf R_{COM} + \mathbf r_i) \times \mathbf p_i = \sum_i \mathbf R_{COM} \times \mathbf p_i+\sum_i \mathbf r_i \times \mathbf p_i = \mathbf R_{COM} \times \sum_i\mathbf p_i+\sum_i \mathbf r_i \times \mathbf p_i$$
As $\sum_i\mathbf p_i = 0$ by hypothesis, $\implies \mathbf L' = \mathbf L$
