Are there any other "involutive" (a la the orthocenter) points on the Euler line?

A point $P$ on the Euler line has barycentric coordinates we can parameterize as $$(a^2 + b^2 - c^2) (a^2 - b^2 + c^2)-p\,(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4)\;:\;\cdots\;:\;\cdots$$ (with the second and third coordinates derived cyclically from the first); here, $p$ is the dilation factor of the circumcenter with respect to the orthocenter. Straightforward symbol-crunching shows that (barring degeneracies in the triangle) the corresponding center of $\triangle PBC$ is $A$ if and only if $p=0$, which makes $P$ the orthocenter.

For the specific symbol-crunching, let the vertices of the triangle have Cartesian coordinates $$A=(0,0) \qquad B = (c,0) \qquad C = (b\cos A, b\sin A)$$ Then $P$ is given by $$\begin{align} x &=\frac{-a^2 + b^2 + c^2 + (a^2-b^2)p}{ 2 c} \\[10pt] y &= \frac{a b ((-a^2+b^2+c^2)(a^2-b^2+c^2) + p(a^4 - 2 a^2 b^2 + b^4 + a^2 c^2 + b^2 c^2 - 2 c^4)}{ 2(-a+b+c)(a+b-c)(a-b+c)(a+b+c)r} \end{align}$$ where $r$ is the circumradius. Then, the corresponding point of $\triangle PBC$ is more of a mess to calculate, since $P$ is already much more complicated than $A$, and since we have to replace $b^2\to|PC|^2$ and $c^2\to|PB|^2$ in the barycentric formulas; luckily, Mathematica doesn't see this as too much of a burden, and dutifully churns-out coordinates of the new point. $$\begin{align} x &= p\;\frac{ \left(\begin{array}{l} \phantom{+} 2 (a^4 b^2 - 2 a^2 b^4 + b^6 + 2 a^2 b^2 c^2 - 2 a^2 c^4 -3 b^2 c^4 + 2 c^6) \\ -p(a^6 + 2 a^4 b^2 - 7 a^2 b^4 + 4 b^6 - 3 a^4 c^2 + 12 a^2 b^2 c^2 - b^4 c^2 - 5 a^2 c^4 - 10 b^2 c^4 + 7 c^6) \\ + p^2(a^6 - 3 a^2 b^4 + 2 b^6 - 2 a^4 c^2 + 6 a^2 b^2 c^2 - b^4 c^2 - 2 a^2 c^4 - 4 b^2 c^4 + 3 c^6) \end{array}\right)}{2 c (-(a^2 + b^2 - c^2) (a^2 - b^2 + c^2) + p(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4))} \\[1em] y &= p\cdot ab\;\frac{\begin{array}{l} \phantom{\cdot}\left( -a^4 + 2 a^2 b^2 - b^4 + 2 b^2 c^2 - c^4 + p(a^2 - b^2 - a c + c^2) (a^2 - b^2 + a c + c^2) \right) \\ \cdot\left( -2 b^2 (a^2 - b^2 + c^2) + p(a^4 + a^2 b^2 - 2 b^4 - 2 a^2 c^2 + b^2 c^2 + c^4) \right) \end{array}}{ 2 (-a + b + c) (a + b - c) (a - b + c) (a + b + c) (\cdots)} \end{align}$$ (Barring degeneracies in the triangle) These coordinates simultaneously vanish (so that this secondary point is the required $A$) when and only when $p$ itself vanishes; that is, whenn $P$ is the orthocenter. $\square$