The symmedian point of the triangle with vertices $A=(-1,0),B=(1,0),C=(x,y)$ is $$F(x,y)=\left(\frac{4 x}{x^2+y^2+3},\frac{2 y}{x^2+y^2+3}\right)$$ By this question $F$ maps $\mathbb{R}^2$ onto an ellipse with foci $A,B$ and eccentricity $\frac{\sqrt3}2$.
Consider the iterated symmedian points $F^2(x,y)=F(F(x,y))$, $F^3(x,y)=F(F(F(x,y)))$, etc.
From the plot, it appears that the iterated symmedian points converge to $(-1,0)$ when starting from left half plane. My attempt to prove this: from the formula, $\frac yx$ gets multiplied by $\frac12$ in each iteration, so the angle $\frac yx$ approaches $0$, so the sequence converge to a point on $x$ axis.
The iterated symmedian points appear to lie on a curve.
What is this curve? Is it an algebraic curve?
Some thoughts:
This curve is defined by Fractional iterates and flows quote Wikipedia:
$f^{(n+m)}(x) = f^{(n)}(f^{(m)}(x))$. This idea can be generalized so that the iteration count $n$ becomes a continuous parameter, a sort of continuous "time" of a continuous orbit.
It looks a bit similar to the power point curve but they are not the same.
The power point curve has barycentric equation $\log(x/y)/\log(y/z)=\log(a/b)/\log(b/c)$.
Then I try to get a barycentric equation for the curve through the iterated symmedian points.
From an exercise on page 90 of Introduction to Triangle Geometry by Paul Yiu:
The distance from $P=(x: y: z)$ to the vertices of triangle $A B C$ are given by \begin{align} A P^2 & =\frac{c^2 y^2+2 S_A y z+b^2 z^2}{(x+y+z)^2}\\ B P^2 & =\frac{a^2 z^2+2 S_B z x+c^2 x^2}{(x+y+z)^2}\\ C P^2 & =\frac{b^2 x^2+2 S_C x y+a^2 y^2}{(x+y+z)^2} \end{align} where $S_A=\frac{b^2+c^2-a^2}{2}$, $S_B=\frac{c^2+a^2-b^2}{2}$, $S_C=\frac{a^2+b^2-c^2}{2}$.
Let $a,b,c$ be the side lengths of $\triangle ABC$.
Then $L=F(C)$ has barycentric coordinates$$(a^2:b^2:c^2)$$
To compute the barycentric coordinates of $F(L)=F^2(C)$,
$$\frac{c^2 \frac{\left(a^2,b^2,c^2\right)}{a^2+b^2+c^2}+|AL|^2 (0,1,0)+|BL|^2 (1,0,0)}{c^2+|AL|^2+|BL|^2}\tag1\label1$$
Using the above formula, I can get $|AL|^2$ in terms of $a,b,c$:
\begin{align}
|AL|^2 & =\frac{c^2 b^4+(b^2+c^2-a^2) b^2 c^2+b^2 c^4}{(a^2+b^2+c^2)^2} \\
&=-\frac{b^2 c^2 \left(a^2-2 b^2-2 c^2\right)}{\left(a^2+b^2+c^2\right)^2}
\end{align}
Similarly $|BL|^2=-\frac{c^2 a^2 \left(b^2-2 a^2-2 c^2\right)}{\left(a^2+b^2+c^2\right)^2}$
Substituting $|AL|^2,|BL|^2$ into \eqref{1}, I get the barycentric coordinates of $F(L)=F^2(C)$ $$\left(3 a^2 \left(a^2+c^2\right),3 b^2 \left(b^2+c^2\right),c^2 \left(a^2+b^2+c^2\right)\right)$$ I verified this formula by GeoGebra.
To find the equation of the curve defined by the iterated symmedian points $C,F(C),F^2(C),\dots$, we need to find an equation satisfied by the barycentric coordinates of $F^n(C)$.
