Let $X$ be a smooth irreducible algebraic variety
over the field of complex numbers ${\mathbb{C}}$.
Let $x\in X({\mathbb{C}})$.
Let $\tau$ be an automorphism of ${\mathbb{C}}$ (not necessarily continuous),
and let $\tau X$ denote the $\tau$-conjugated ${\mathbb{C}}$-variety obtained from $X$ by transport of structure
(i.e. by action of $\tau$ on the coefficients of equations defining $X$).
We consider the *topological* fundamental groups $\pi_1(X({\mathbb{C}}),x)$ and $\pi_1((\tau X)({\mathbb{C}}),\tau x)$.

In the papers of Serre, Exemples de variétés projectives conjuguées non homéomorphes,
C. R. Acad. Sci. Paris **258** (1964), 4194–4196,
and of Milne and Suh,
Nonhomeomorphic conjugates of connected Shimura varieties,
one can find examples of $X$ and $\tau$ such that
$\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are *not* isomorphic.
The authors conclude that in these cases the topological spaces $(\tau X)({\mathbb{C}})$ and $X({\mathbb{C}})$ are not homeomorphic.

In my very recent preprint with Cyril Demarche
(excuse me for advertising my own work!) we consider the following situation.
Let $X=G/H$, where $G$ is a connected linear algebraic group over ${\mathbb{C}}$,
and $H\subset G$ any algebraic subgroup, *not necessarily connected*.
Set $x:=eH\in X({\mathbb{C}})$.
We prove that in this case $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are canonically isomorphic.
I am trying to understand, what this really means.

Question.For a homogeneous space $X=G/H$ over ${\mathbb{C}}$ as above, and for $\tau\in {\rm Aut}({\mathbb{C}})$, is it always true that

(1) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homotopically equivalent, or even

(2) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homeomorphic, or even

(3) $\tau X$ and $X$ are isomorphic ${\mathbb{C}}$-varieties?