Are there any other "involutive" (a la the orthocenter) points on the Euler line? Throughout, given a triangle $T$ let $G(T)$ and $H(T)$ be the centroid and orthocenter of $T$, respectively. For $r\in\mathbb{R}$ and $p\in\mathbb{R}^2$, let $h_{p:r}$ be the homothety with focus $p$ and factor $r$ (with the understanding that $h_{p:0}(q)=p$ for all $q\in\mathbb{R}^2$). So, for example, the Euler line of a triangle $T$ is $$\{h_{G(T): r}(H(T)): r\in\mathbb{R}\}.$$
I've asked before about "generalized triangle centers" satisfying the same involutive property as the orthocenter. This question is a much more local version of that one: roughly speaking, are there any such generalized triangle centers (besides the orthocenter itself) which always lie on the Euler line?

Precisely, say that an Eulerian triangle center function is a continuous function $f:D\rightarrow\mathbb{R}$ with the following properties:

*

*$D$ is a dense open connected subset of $(\mathbb{R}^2)^3$.


*Both $D$ and $f$ are $S_3$-invariant: for each permutation $\sigma\in S_3$ and $(p_1,p_2,p_3)\in D$ then $(p_{\sigma(1)},p_{\sigma(2)},p_{\sigma(3)})\in D$ and $f(p_1,p_2,p_3)=f(p_{\sigma(1)},p_{\sigma(2)},p_{\sigma(3)})$.


*Both $D$ and $f$ are similarity-invariant: if $\triangle pqr$ is similar to $\triangle p'q'r'$ and $(p,q,r)\in D$, then $(p',q',r')\in D$ and $f(p,q,r)=f(p',q',r')$.
If $f$ is an Eulerian triangle center function, let $$\hat{f}:D\rightarrow \mathbb{R}^2: (p,q,r)\mapsto h_{G(\triangle pqr): f(p,q,r)}(H(\triangle pqr))$$ be the corresponding appropriately-scaled point on the Euler line. So, for example, the constant map ${\bf 1}:(p,q,r)\mapsto 1$ recovers the orthocenter $\hat{\bf 1}: (p,q,r)\mapsto H(\triangle p,q,r)$ (modulo issues re: degenerate triangles).
My question is:

Is there a non-constantly-$1$ Eulerian triangle center $f:D\rightarrow\mathbb{R}$ such that, for comeager-many triples of points $(p,q,r)\in(\mathbb{R}^2)^3$, we have $$\hat{f}(p, q, \hat{f}(p,q,r))=r$$ (so basically, it's not the orthocenter but it satisfies the above-mentioned involutive property)?

If the answer is positive, my next question is how many there are - up to the obvious equivalence relation of "agree on a dense subset of the intersection of their domains." I tentatively suspect the answer is negative, however.

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. $\square$
