Let $A_1A_2\cdots A_{k+3}$ be a polygon inscribed in a circle for $k\geq 0$ and $X$ another point on the circle. Call $A_1A_2\cdots A_{k+3}$ the zero-polygon. And derive the $(l+1)$-polygon from $l$-polygon as follows:
Let $P_1P_2\cdots P_{k+3}$ be the $l$-polygon and $Q_i$ the projection of $X$ on line $P_i P_{i+1}$ for $1\leq i\leq k+3$, assuming that $P_{k+4}=P_1$. Then $Q_1Q_2\cdots Q_{k+3}$ is the $(l+1)$-polygon.
I discover that the $(k+1)$-polygon degenerates to a line whereas, for each $1\leq i\leq k$, the $i$-polygon does not. When $k=0$, this is the well-known Simson-Wallace theorem.
I am able to prove the $k=1$ case as follows: it is equivalent to that $X$ is the Miquel point of the quadrilaterial $B_1B_2B_3B_4$. Extend $B_2B_3$ to intersect with $B_1B_4$ at $P$, then we only need that $XB_1B_2P$ lie on the same cycle. Noted that $XB_1\perp A_2B_1,XB_2\perp A_2B_2$, we will prove that $A_2B_1B_2P$ lie on the same cycle. By the time this is proved, we will know that the pentagon $XB_1A_2PB_2$ is cyclic. In a word, we need $\angle A_2B_1B_4=\angle A_2B_2P=\angle B_3B_2A_3$. This can be proved by juggling with the angles (believe me that I have checked all details).
The $k=2$ case will be simple once we use the $k=1$ case to help and in a similar fashion. (Yes, I have written a proof as well, but too tedious to present it here.)
This result has been checked for $k=1,\cdots,7$ using my computer.
I am seeking for a prove of the general case, which seemingly should be an algebraic one. It would be marvellous if someone managed to prove it using geometric methods.
Any related discussion is welcomed! THX:)
One can show that the projection from $Z$ to the line through $Z-2P$ and $Z-2Q$ (for $Z$, $P$, $Q$ complex numbers) is given by $$Z' = Z - \frac{\overline{P}Q - P \overline{Q}}{\overline{P} -\overline{Q}} \tag{$\star$}$$ Defining $[x]:=\frac12(1-x)$, if $$P = [P'] M \qquad Q = [Q'] M$$ for some $P'$ and $Q'$ on the unit circle, and $M$ some arbitrary complex number, then we can compute $$Z' = Z - 2\; [P']\;[Q'] \; M \tag{$\star\star$}$$ If, in particular, $Z=1+0i$, this says
Interpreting this result geometrically is left as an exercise to the reader.
Now, to prove OP's conjecture ...
Changing OP's notation, consider point (and complex number) $X$ and the $n$ points/numbers $Y_0$, $Y_1$, $\ldots$, $Y_{n-1}$ on the unit circle; without loss of generality, we can take $X = 1+0i$.
Let $Y_k^{(1)}$ (where the $1$ is not a power, but an index) be the foot of the perpendicular from $X$ to line $Y_k Y_{k+1}$. By $(\star\star\star)$ with $M=1$, $P'=Y_k$, $Q'=Y_{k+1}$, we have $$[Y_k^{(1)}] = [Y_k][Y_{k+1}]\tag1$$ Let $Y_k^{(2)}$ be the foot of the perpendicular from $X$ to line $Y_k^{(1)} Y_{k+1}^{(1)}$. We cannot simply write $[Y_k^{(2)}]=[Y_k^{(1)}][Y_{k+1}^{(1)}]$ as above, since $Y_k^{(1)}$ and $Y_{k+1}^{(1)}$ are not (necessarily) on the unit circle. However, unpacking equation $(1)$, and taking $M=[Y_{k+1}]$, we have $$Y_k^{(1)} = 1 - 2 [Y_k][Y_{k+1}] = 1-2[Y_k]M \qquad Y_{k+1}^{(1)} = 1-2[Y_{k+1}][Y_{k+2}] = 1-2[Y_{k+2}]M$$ so that $$[Y_k^{(2)}] = [Y_k][Y_{k+1}]M = [Y_k][Y_{k+1}][Y_{k+2}] \tag2$$ Likewise, taking $M=[Y_{k+1}][Y_{k+2}]$, then $M=[Y_{k+1}][Y_{k+2}][Y_{k+3}]$, etc, we find $$\begin{align} [Y_k^{(3)}] &= [Y_k][Y_{k+1}]\cdots[Y_{k+3}] \\ [Y_k^{(4)}] &= [Y_k][Y_{k+1}]\cdots[Y_{k+4}] \\ \cdots\; &= \;\cdots \\ [Y_k^{(n-1)}] &= [Y_k][Y_{k+1}]\cdots[Y_{k+n-1}] \end{align} \tag{3}$$ Note that the last of these, $[Y_k^{(n-1)}]$, is simply $[Y_0][Y_1]\cdots[Y_{n-1}]$ for all $k$, which implies that $Y_0^{(n-1)}$, $Y_1^{(n-1)}$, $\ldots$, $Y_{n-1}^{(n-1)}$ are all the same point. As this point is the foot of the perpendiculars from $X$ to each of the lines $Y_0^{(n-2)}Y_1^{(n-2)}$, $Y_1^{(n-2)}Y_2^{(n-2)}$, $\ldots$, those lines must coincide; ie, the points $Y_0^{(n-2)}$, $Y_1^{(n-2)}$, $Y_2^{(n-2)}$, $\ldots$ are collinear. $\square$