What does it mean mathematically the position of the center of mass relative to other particles is the same? My understanding is that the center of mass of a system of particles is that an imaginary point that travels in space and has position vector $$\vec r = \frac{\sum_{i=1}^{n}m_i\vec r_i}{\sum_{i=1}^nm_i}.$$
My textbook suggests that if you use one coordinate system to find the center of mass of an object, its position relative the particles is the same is the object moves or the coordinate system changes.
What does it mean mathematically that the position of the center of mass relative to the positions of the particles is the same? I found if there is some rotation, its position vectos relative to all the other particles change, like the object below.

 A: The statement you are asking about is the following, I think.
Let's say we compare the positions measured in two different coordinate systems $A$ and $B$, i.e. $\vec{r}_{i,A},\,\vec{r}_A$ and $\vec{r}_{i,B},\,\vec{r}_B$. Now let's further say that coordinates in the two different coordinate systems are related by one of the isometries of Euclidean space, i.e.
$$
\vec{r}_{i,A} = M \vec{r}_{i,B} + \vec{a}
$$
with an orthogonal transformation $M$ and a translation $\vec{a}$.
You will then find that for the center of mass
$$
\vec{r}_{A} = M \vec{r}_{B} + \vec{a}
$$
and consequently
$$
\vec{r}_{A} - \vec{r}_{i,A} = M \left(\vec{r}_{B} - \vec{r}_{i,B}\right)\,,
$$
i.e. the relative distance between a particle and the center of mass of the system is not affected by the translation, it is only rotated and its absolute value is unchanged by the transformation between the two coordinate systems.
A: 
"My understanding is that the center of mass of a system of particles is that an imaginary point that travels in space and has position vector..."

When we look at the image you posted and take the example of the three individual masses, we can see that the center of mass (COM) exists in an empty region between the masses. This "imaginary" point travels with the masses if they physically move, and remains constant relative to the masses so long as the orientation of the masses stays the same.
Where I think you have confusion is in the choice of reference when we calculate the location of the COM. Vector $\vec{r}_{com}$ only exists relative to whatever point you choose to calculate it from, and all it tells you is the position of the COM relative to that chosen point. You can choose any point, but $m_1$ is conveniently located at the origin of this coordinate system, so it is logical to choose it as the starting point of $\vec{r}_{com}$. But you could easily have a different $\vec{r}_{com}$ if you were to choose $m_2$ or $m_3$ as your starting point. You can visually see that choosing $m_3$ instead of $m_1$ to start from would result in a shorter $\vec{r}_{com}$ and pointing downwards.

"My textbook suggests that if you use one coordinate system to find the center of mass of an object, its position relative the particles is the same is the object moves or the coordinate system changes."

As discussed above, you can choose the coordinate system for calculation of COM position based on what makes sense for the situation. So if we wanted to find $\vec{r}_{com}$ from $m_3$, maybe it would be easier to choose a coordinate system where $m_3$ is the origin.
Regardless of where you choose to measure the COM from, the physical location of the COM only depends on the positions of the masses themselves.

"What does it mean mathematically that the position of the center of mass relative to the positions of the particles is the same? I found if there is some rotation, its position vectos relative to all the other vectors change, like the object below."

Mathematically, this is saying that the physical distance between each mass $m_1,m_2,m_3$ individually, and the COM, remains constant regardless of the movement of the masses if and only if all the masses move in the exact same way at the same time.
If you imagine the triangular formation of the masses to be 'fixed' (in that when one moves, they all move by the same amount) then the COM remains fixed relative to the masses that contribute to it. So to put this into perspective, if all three masses were translated together upwards by 10 units, the COM also moves up by the same 10 units, but the COM would physically be the same distance away from $m_1, m_2$ & $m_3$ individually as they were before the translation. Same goes for any rotation. Imagining that the masses all rotate about $m_2$ as a rotation axis, you'll see the COM move with the masses, preserving the individual distances between the COM and the masses. Now it is up to you if the COM moves, where you want to then calculate your $\vec{r}_{com}$ from within your chosen coordinate system.
Now, and finally, if any of the masses were to move independently from the others, this would entirely change the COM of the system, and consequently the $\vec{r}_{com}$ if our origin is the same.
Take the scenario where we only move mass $m_3$ upwards by 10 units, and the other masses stay where they are. The COM is now in a different position relative to the origin, and each $\vec{r}_{com}$ from each mass to the COM is different.
A: look at this figure

The vector u is the position between the center of mass and the particle position.
you can describe this vector in a body fixed coordinate system that located at the center of mass ( blue one). Now if you use other intertial system $I_2$ the center of mass vector and the particle position changes, but the vector u doesn’t change in body fixed coordinate system.
If the center of mass coordinate system (blue one) is moving, vector u is remaining unchanged in the body coordinate system
