The answer is "yes" (even if one excludes the elementary cases depicted in my first answer, and by a certain number of us).
Let $A_k$ ($k=0,\cdots 6$) be the vertices and $P_k$ the points on line segment $[A_k,A_{k+1}]$ (with a modulo 7 convention).
We have:
$$\underbrace{\frac17 \sum \vec{OP_k}}_{\text{c. of mass of points} \ P_k}=\underbrace{\frac17 \sum \vec{OA_k}}_{\text{c. of mass of points} \ A_k \ = \ O}+\frac17 \sum \vec{A_kP_k} \tag{1}$$
Let
$$v_k:=\vec{A_kP_k} \ \ \text{and} \ \ \vec{V}= \sum \vec{A_kP_k} \tag{2}$$
As a consequence of (1); the two centers of mass coincide if and only if
$$\vec{V}=0 \tag{3}$$
This is possible in a myriad of cases, as the following graphical representation shows. Indeed (3) is possible if and only, with the $v_k$s, one can built an heptagon of a special kind (irregular sidelengths and regular angles; i.e., all its internal angles are equal) with a "head-to-tail" loop representation (like in statics).
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/fe0X4.jpg)
Such heptagons can be obtained graphically as the intersection of strips with variable width (or, if one prefers, of half-spaces) having polar angles $k \dfrac{2 \pi}{7}$. Slightly moving these strips in a orthogonal direction give different heptagons, i.e., different set of points $P_k$.
Now that we have understood the idea, a rigorous construction of any case is made possible by incremental construction. Here is how:
Let us assume that points $P_k$ are referenced by their abscissas $a_k$ defined by $\vec{A_kP_k}=a_k \vec{A_kA_{k+1}}.$ We will also consider that the sides have unit length.
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- Start with any case for values $a_k$ (for example all $a_k=1/2$ (midpoints)).
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- Make a choice of a certain index $k$ and a certain (small) value $d$.
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- change $a_k$ into $a_k+2d$ (the abscissa is either augmented of shortened according to the sign of $d$).
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- change $a_{k-1}$ into $a_{k-1}-d/\cos (2\pi/7)$.
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- change $a_{k+1}$ into $a_{k+1}-d/\cos (2\pi/7)$.
Then iterate the process 1)+2)+3)+4) (of course one has to check that abscissas always remain in the range $[0,1]$).
Remarks :
Angle $2\pi/7$ above is the so-called "external angle" of the heptagon.
A similar question could have been asked for any regular $n$-gon.
A question of the same type related to a general triangle (not equilateral in general): Coinciding centroids of two triangles.
As remarked by @dxiv, this center of mass preservation is not limited to the case where this center is $O$. It can be applied to any center of mass.
I end with a simulation result showing where centers of mass of points $P_k$ can be situated for random positions of these points on the sides (purple points) and for extreme cases where points $P_k$ are situated on some $A_k$s (red points).
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/8iLl8.jpg)
Remark: A cousin issue, asked for a pentagon instead of a heptagon, is to be found here.