An order-$m$ subplane of a finite projective plane of order $n$ is called a *Baer subplane* if $n=m^2$. 


It is known that the projective plane $PG(2,q)$ is a Baer subplane of the Desarguesian projective plane $PG(2,q^2)$. 


**Question**: Is it the only Baer subplane? In other words, are all Baer subplanes of $PG(2,q^2)$ isomorphic up to automorphisms of $PG(2,q^2)$?



Is there a unique Baer subplane in a finite Desarguesian projective plane?
The answer is true.


Facts:


Any quadrangle can be mapped onto any other quadrangle by a collineation in a Desarguesian plane. See [this paper](https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/jlms/s1-33.1.25).


Every Baer subplane contains a quadrangle.


Every quadrangle lies on a unique Baer subplane in $PG(2,q^2)$. See [this paper](https://link.springer.com/article/10.1007/s10623-011-9567-0).


So if $A$ and $B$ are two Baer subplanes, let $a$ be a quadrangle in $A$, and $b$ a quadrangle in $B$. Then there's an automorphism $φ$ of $PG(2,q^2)$ mapping $a$ to $b$. It follows that $φ(A)=B$ by uniqueness. 


Therefore, all Baer subplanes of $PG(2,q^2)$ are isomorphic up to automorphisms of $PG(2,q^2)$.


