TITLE: Are there infinitely many "generalized triangle vertices"?
QUESTION [21 upvotes]: Briefly, I'd like to know whether there are infinitely many "generalized triangle centers" which - like the orthocenter - are indistinguishable from a vertex of the original triangle. This is basically a refinement of this MSE question of mine; note that the type of "generalized triangle center" I'm interested in is not the standard one (see 1,2), although I would also be interested in the situation for that definition.
Separately, since MO bounties have time limits, I'll "informally bounty" this question: I'll reward the first complete answer to the question with a 1000 point bounty, whenever - if ever - that should happen.

Definitions
Let $\mathbb{T}$ be the set of noncollinear ordered triples of points in $\mathbb{R}^2$. Say that a topological triangle center representative (ttcr) is a function $t:G\rightarrow \mathbb{R}^2$ such that:

$G$ is a connected dense open subset of $\mathbb{T}$ and $t$ is continuous;

$G$ and $t$ are each symmetric: if $(a,b,c)\in G$, then $(a,c,b)$ and $(b,a,c)$ are in $G$ as well and we have $t(a,b,c)=t(a,c,b)=t(b,a,c)$;

both $G$ and $t$ are homothety-etc.-invariant: if $\alpha:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is a composition of rotations, reflections, translations, and homotheties, and $(a,b,c)\in G$, then $(\alpha(a),\alpha(b),\alpha(c))\in G$ and $t(\alpha(a),\alpha(b),\alpha(c))=\alpha(t(a,b,c))$.

and $t$ is (usually) iterable: for a dense open $H\subseteq G$, if $(a,b,c)\in H$ then $(t(a,b,c),b,c)\in H$.


Each classical triangle center that I'm familiar with corresponds to a ttcr, possibly after tweaking the domain. For instance, in the case of the orthocenter we need to throw out right triangles to satisfy the iterability requirement.
A topological triangle center is then an equivalence class of ttcrs with respect to the relation $t\sim s\iff t_{\upharpoonright \operatorname{dom}(t) \cap \operatorname{dom}(s)}=s_{\upharpoonright \operatorname{dom}(t)\cap \operatorname{dom}(s)}$. Finally, a pseudovertex is a topological triangle center with a representative $t$ satisfying $$t(t(a,b,c),b,c)=a$$ for every $(a,b,c)\in \operatorname{dom}(t)$.

Question
My question is simply, how many pseudovertices are there? Specifically:

Are there infinitely many pseudovertices?

I strongly suspect that the answer is yes (indeed that there should be continuum-many due to the existence of at least one not-too-interesting continuously-parameterized family), and I suspect that in fact there is an easy proof of this fact, but I can't see it at the moment.
So far I know of three distinct pseudovertices (modulo appropriate definitional abuse):

The orthocenter, $X(4)$.

The isogonal conjugate of the Euler infinity point, $X(74)$.

The isogonal conjugate of Parry's reflection point, $X(1138)$.


As a curiosity, note that these three centers are nontrivially related to each other: it turns out that $X(74)$ is the crosspoint of $X(4)$ and $X(1138)$. This fact, as well as the examples of $X(74)$ and $X(1138)$, was found by MSE user Blue at the above-linked question.

REPLY [3 votes]: Transferring the Addendum to my previous non-answer to a non-answer of its own, as it's beginning to sprawl (and because it actually seems more significant) ...

It happens that $X(4)$, $X(74)$, $X(1138)$ also lie on the circumcubic $K279$, for which there is this geometric description attributed to Angel Montesdeoca:

Let $\triangle A' B' C'$ be the cevian triangle of a point $P$, let $A^\star$, $B^\star$, $C^\star$ be the circumcenters of $\triangle AB'C'$, $\triangle A'BC'$, $\triangle A'B'C$, and let $Q$ be the circumcenter of $\triangle A^\star B^\star C^\star$. Then $K279$ is the locus of $P$ such that $Q$ lies on the Euler line.

Unlike with Neuberg, given a point $P$ on this curve, the corresponding cubic for $\triangle PBC$ does not necessarily pass through $A$. Indeed (barring triangle degeneracies), Mathematica-assisted symbol crunching (see below) shows that this occurs for (Kimberling-esque) triangle center $P$ when and only when $P$ is one of $X(4)$, $X(74)$, $X(1138)$. Effectively, then, the property

$P$ lies on the $K279$ of $\triangle ABC$, and $\triangle ABC$ is inscribed in the $K279$s of $\triangle PBC$, $\triangle APC$, $\triangle ABP$.

characterizes the known pseudovertices. This could be useful in the search for (or the ruling-out of?) other candidates.

For the symbol-crunching, we start with the barycentric $u:v:w$ equation for $K279$ relative to $\triangle ABC$:
$$\sum_{cyc} b^2 c^2 ((b^2 - c^2)^2 + a^2 (b^2 + c^2 - 2 a^2))\; u (v^2 - w^2) = 0 \tag{1}$$
Then, considering $P$ with barycentric coordinates $u:v:w$, the $K279$ with respect to $\triangle PBC$ has barycentric $u':v':w'$ equation that derives from $(1)$ by substituting $u\to u'$, $v\to v'$, $w\to w'$, and
$$\begin{align}
b^2 &\to |PC|^2 = \frac{a^2 v^2 + b^2 u^2 + (a^2+b^2-c^2) u v}{(u + v + w)^2} \\[0.5em]
c^2 &\to |PB|^2 = \frac{a^2w^2 + c^2 u^2 + (a^2-b^2+c^2) u w }{(u+v+w)^2}
\end{align}$$
Since $A$ has barycentric coordinates $u':v':w' = u+v+w:-v:-w$ relative to $\triangle PBC$, we substitute those, as well. This yields a degree-7 $uvw$ polynomial, whose $184$ terms I won't transcribe here. Mathematica confirms that the polynomial vanishes if $P$ is one of $X(4)$, $X(74)$, $X(1138)$. (It does not vanish when $P$ is $X(2)$, the centroid of $\triangle ABC$, unless $b=c$.)
Applying Mathematica's Resultant to eliminate $u$ from this and $(1)$ gives this equation (ignoring powers of factors):
$$vw(v + w)\cdot(v-w)\cdot f_4 \cdot f_{74} \cdot f_{1138} \cdot f = 0$$
Here,
$$f_4 := v(-b^2+c^2+a^2) -w(-c^2+a^2+b^2)$$
so that
$$f_4 = 0 \quad\implies\quad v = \frac{k}{-b^2+c^2+a^2} \quad w = \frac{k}{-c^2+a^2+b^2} \quad\left(u = \frac{k}{-a^2+b^2+c^2}\right)$$
(with the $u$-coordinate symmetrically completing the "center"). Thus, this factor can vanish only at center $X(4)$. Likewise, $f_{74}$ and $f_{1138}$ are linear factors in $v$ and $w$ that can vanish only at centers $X(74)$ and $X(1138)$.
Further, our interest in centers allows us to ignore factors $v$, $w$, $v+w$. Factor $v-w$ vanishes when $v=w(=u)$, which corresponds to centroid $X(2)$, a candidate we have already ruled-out.
Factor $f$ is a degree-$6$ polynomial in $v$ and $w$, with $711$ terms. I haven't done a proper analysis of it; however, when, say, $(a,b,c)=(6,9,13)$, the solutions are all non-real. I'm pretty sure we can consider this extraneous in general. (There's probably a neat way to avoid this factor altogether by first reducing the system or something, but I haven't found it.)
Of course, it would be best to devise a more-direct, less-computational argument, perhaps one based on Montesdeoca's geometric description of $K279$. (One may note that the cevian triangle of $P$ with respect to $\triangle ABC$, and that of $A$ with respect to $\triangle PBC$, are always the same triangle, so there's a start! :) That's what I'm investigating now.