TITLE: Conjugation of homogeneous spaces
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: The result in my preprint mentioned in the question was erroneous (the mistake was noticed by a referee). It is possible to construct a quotient  $X=G/H$ and and automorphism $\tau$ of $\mathbb{C}$ such that $\pi_1(\tau X(\mathbb{C}))$ is not isomorphic to $\pi_1(X(\mathbb{C}))$, see my preprint with Yves Cornulier based on his answer to this question. In our example  $G={\rm SL}(n,\mathbb{C})\times \mathbb{C}^*$ with $n\ge 5$, and $H$ is a finite nonabelian subgroup of order 55.
This time the referee noticed no mistakes... Thus the answer to the questions (1), (2), and (3) is NO.