This answer has an arithmetic error and now is under revision.
We can consider all given points as complex numbers, points of the complex plane. Then, as far as I understood, for all integers $m\ge 1$, $1\le j\le k$ we we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2},$$
where $\xi=\exp\frac{2\pi i}{k}$.
Since $A_{m+1,1}= A_{m,3}$, we have
$$A_{m+1}+(P_{m+1}-A_{m+1})\xi^{-1}=A_m+(P_m-A_m)\xi.$$
It follows
$$A_{m+1}=-A_m\xi+\frac{P_m\xi^2-P_{m+1}}{\xi-1}.$$
This formula suggests that the chain can be closed for a specific choice of $O$ and $P_m$’s, but I guess that you are looking for a stable general pattern. So let’s look when the sequence $A_n$$\{A_m\}$ is periodic.
Putting $B_m=A_m(-\xi)^m$$B_m=A_m(-\xi)^{-m}$, we obtain
$$B_{m+1}=B_m+\frac{P_m\xi^2-P_{m+1}}{\xi-1}(-\xi)^{m-1}.$$$$B_{m+1}=B_m+\frac{P_m\xi^2-P_{m+1}}{\xi-1}(-\xi)^{-m-1}.$$
Since the sequence $\{P_n\}$ has a period $n$, we have
$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^n,$$$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^{-n},$$
that is
$$A_{m+2n}-A_{m+n}(1+\xi^{-n})+A_m=0.$$$$A_{m+2n}-A_{m+n}(1+(-\xi)^{n})+A_m(-\xi)^{n}=0.$$
An equation $\lambda^2-(1+(-\xi)^{n})\lambda +(-\xi)^{n}$ has roots $1$ and $(-\xi)^{n}$. The following cases are possible.
1)) $(-\xi)^{-n}=1$. This holds iff $\xi^{-n}=1$, that($n$ is even and $k|n$. Then) or $A_{m+2n}-2A_{m+n}+A_m=0$($n$ is odd, that$k$ is even and $k|2n$). The theory of recurrence relations implies that $A_{r n+m}=c_1(m) + c_2(m)r$ for each $r$ and some constants $c_1(m)$ and $c_2(m)$ depending on $m$. If all $c_2(m)$ are zeroes then the sequence $\{A_m\}$ has a period $n$ (or its divisor). Otherwise the sequence $\{A_{m+rn}\}_{r\in\Bbb N}$$\{A_m\}$ is an arithmetic progressionnot periodic. ItThus the sequence $\{A_m\}$ is periodic iff for each $m$ we have $A_m=A_{m+n}$. SinceThis can happen iff the choice of $k|n$$P_m$’s is specific. Namely, we have
$$0=A_{m+n}(-\xi)^{m+n}- A_{m}(-\xi)^{m}=B_{m+n}-B_{m}=$$ $$\frac {(-\xi)^{m-1}}{\xi-1}\sum_{j=1}^n - P_{m+j} (-\xi)^{j-1}(\xi^3+1)=$$ $$-\frac {(-\xi)^{m-1}(\xi^3+1)}{\xi-1}\sum_{j=1}^n P_{m+j} (-\xi)^{j-1}.$$ 1.1)) If $n$ is even and $k|n$ then
$$\sum_{j=1}^n P_{m+j} (-\xi)^{j-1}=0,$$$$0=A_{m+n}(-\xi)^{-m-n}- A_{m}(-\xi)^{-m}=B_{m+n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^n P_{m+j} (-\xi)^{-(m+j)},$$
that is, $-\xi$$(-\xi)^{-1}$ is a root of a polynomial $P(x)=\sum_{j=1}^n P_{j} x^j$. It follows that
1.2)) If $A_{m+n}=A_m$ for each$n$ is odd, $m$$k$ is even, and $k|2n$ then
$$0=A_{m+2n}(-\xi)^{-m-2n}- A_{m}(-\xi)^{-m}=B_{m+2n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^{2n} P_{m+j} (-\xi)^{-(m+j)},$$
that is, the the sequence $\{A_m\}$ has period$(-\xi)^{-1}$ is a root of a polynomial $n$.$(1+x^n)P(x)=\sum_{j=1}^{2n} P_{j} x^j$
2)) $\xi^{-n}\ne 1$, that is $k\not|n$$(-\xi)^{-n}\ne 1$. It is easy to check that an equation $\lambda^2-(1+\xi^{-n})\lambda +1$ has two distinct roots(This case holds, in particular, when both $\lambda_1,\lambda_2$$n$ and $k$ are odd). The theory of recurrence relations implies that $A_{r n+m}=C_1\lambda_1^r+ C_2\lambda_2^r$$A_{r n+m}=c_1(m) + c_2(m)(-\xi)^{nr}$ for each $r$ and some constants $C_1$$c_1(m)$ and $C_2$. Since $\lambda_1\lambda_2=1$, if not $|\lambda_1|=|\lambda_2|$ and the sequence $A_{rn+m}$ is non-zero then$c_2(m)$ depending on $|A_{rn+m}|$ either grows to infinity or vanishes to zero. It both cases the sequence $A_{rn+m}$ is not periodic$m$. If all $|\lambda_1|=|\lambda_2|$$c_2(m)$ are zeroes then, since $\lambda_1\lambda_2=1$, we have $\lambda_2=\overline{\lambda_1}$, and so the sequence $1+\xi^{-n}=-(\lambda_1+\lambda_2)$ is$\{A_m\}$ has a real numberperiod $n$ (or its divisor). ThusOtherwise $\xi^{-n}$$-\xi$ is a real number distinct fromprimitive $1$, that is $\xi^{-n}=-1$$q$-th root of unity, where $$q=\cases{k, \mbox{ if }k\equiv 0\pmod 4\\ k/2, \mbox{ if }k\equiv 2\pmod 4\\ 2k, \mbox{ if }k\equiv 1,3\pmod 4}.$$
Remark that is $n\equiv k/2 \pmod k$ $q=\frac{2k}{\gcd(k-2,2k)}=\frac{2k}{\gcd(k-2,4)}$. ThenThus $\lambda_{1,2}=\pm 1$$(-\xi)^n$ is a primitive $\tfrac{q}{\gcd(q,n)}$-th root of unity, and so the sequence $A_{rn+m}=C_1+ C_2\cdot (-1)^r$$\{A_m\}$ has a period $\tfrac{qn}{\gcd(q,n)}=\operatorname{lcm}(q,n)$ (or its divisor). Moreover, for each $r$ and some constants $C_1$ and$m$, points $C_2$. It follows$\{A_{r n+m}: 0\le r\le q-1\}$ are vertices of a $A_m=A_{m+2n}$$q$-qon. Since $k|2n$, we have
$$0=A_{m+2n}(-\xi)^{m+2n}- A_{m}(-\xi)^{m}=B_{m+n}-B_{m}=$$ $$\frac {(-\xi)^{m-1}}{\xi-1}\sum_{j=1}^{2n} - P_{m+j} (-\xi)^{j-1}(\xi^3+1)=$$ $$-\frac {(-\xi)^{m-1}(\xi^3+1)}{\xi-1}\sum_{j=1}^{2n} P_{m+j} (-\xi)^{j-1}.$$
$$\sum_{j=1}^{2n} P_{m+j} (-\xi)^{j-1}=0,$$ that is Finally, recall that for each $m\ge 1$, $-\xi$ is a root of a polynomial$1\le j\le k$ we have $(1+x^n)P(x)=\sum_{j=1}^{2n} P_{j} x^j$$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2}$. It follows that if the sequence $A_{m+2n}=A_m$$\{A_m\}$ has a period $p$ then for each fixed $m$, that is, the$j$ a sequence $\{A_m\}$$\{A_{m,j}\}$ has a period $n$$\operatorname{lcm}(p,n)$ (or its divisor).