Is every maximal $p$-subgroup of $\operatorname{PGL}(2,K)$ conjugate, where $p$ is an odd prime not equal to the characteristic of $K$?
Here $\operatorname{PGL}(2,K)$ is the quotient group of the group of 2×2 invertible matrices with entries from the field $K$ modulo the group of nonzero $K$-scalar of multiplies the identity matrix.
The only $p$-subgroups are isomorphic to the locally cyclic groups of $p^n$th roots of unity in $F$, where $[F :K]≤2$, that is, $F$ is a quadratic field extension of $K$ or $F = K$. Assuming the $p$-subgroup generates $F$ over $K$ and is not the identity when $F = K$, the normalizers are (generalized) dihedral groups, \operatorname{Dih}(F^×), which proves the maximality of the full groups of roots of $p^n$th roots of unity in $F$.
In the finite field case, things are easier. F^×/K^× has order $q+1$ and K^× has order $q−1$, so either $p$ divides the first, or $p$ divides the second, and thus all $p$-subgroups of the same order are conjugate just by bringing their torus into line. This is not surprising since the maximal $p$-subgroups are conjugate by Sylow's theorem, and cyclic, so have a single subgroup of each order.
In the infinite field case, I am not quite as certain. For one thing, the quadratic $F$ is no longer uniquely defined by $K$, so I'm not sure to what extent the non-split tori are conjugate. Even worse, I'm not even sure that one cannot have a maximal $p$-subgroup of $K$-type and a maximal $p$-subgroup of $F$-type.