Call $[A_n,A_n+V_n]$ the segment at time $n$ assuming that $A_n$ is the end of the segment around which it rotates between times $n$ and $n+1$. Thus $A_n$ and $V_n$ are complex numbers, $A_0=1=V_0$, and $V_n$ has modulus $1$ for every $n$. Then (I believe that) for every $n\ge0$, $$A_{n+1}=A_n+U_{n+1}V_n,\qquad V_{n+1}=-U_{n+1}V_n,$$ where $(U_n)_{n\ge0}$ is an i.i.d. sequence with a given distribution on the unit circle.
This shows that $(A_n)_n$ is not a Markov chain in general except for the case you said you were interested in, where the random variables $U_n$ are uniform on the unit circle, that $(A_n,V_n)_n$ is always a Markov chain, and that $$V_n=(-U_n)\cdots(-U_1)(-U_0),\qquad (-A_n)=\sum_{k=1}^n(-U_1)\cdots(-U_k). V_n=(-U_1)\cdots(-U_n),\qquad A_n=1-\sum_{k=1}^n(-U_1)\cdots(-U_k).$$ We seem to be back to some known territory here...
Call $[A_n,A_n+V_n]$ the segment at time $n$ assuming that $A_n$ is the end of the segment around which it rotates between times $n$ and $n+1$. Thus $A_n$ and $V_n$ are complex numbers, $A_0=1=V_0$, and $V_n$ has modulus $1$ for every $n$. Then (I believe that) for every $n\ge0$, $$A_{n+1}=A_n+U_{n+1}V_n,\qquad V_{n+1}=-U_{n+1}V_n,$$ where $(U_n)_{n\ge0}$ is an i.i.d. sequence with a given distribution on the unit circle.
This shows that $(A_n)_n$ is not a Markov chain in general except for the case you said you were interested in, where the random variables $U_n$ are uniform on the unit circle, that $(A_n,V_n)_n$ is always a Markov chain, and that $$V_n=(-U_n)\cdots(-U_1)(-U_0),\qquad (-A_n)=\sum_{k=1}^n(-U_1)\cdots(-U_k).$$ We seem to be back to some known territory here...