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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 $PG(2,q^2)$?

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The answer is true.

Facts:

Any quadrangle can be mapped onto any other quadrangle by a collineation in a Desarguesian plane. See this paper.

Every Baer subplane contains a quadrangle.

Every quadrangle lies on a unique Baer subplane in $PG(2,q^2)$. See this paper.

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 $PG(2,q^2)$.

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