If $B$ is a Borel subgroup of connected linear algebraic group $G$, the fact that all Borel subgroups of $G$ are conjugate, combined with the fact that $N_G(B) = B$, shows that the set of Borel subgroups of $G$ can be identified with the set of left cosets of $B$ in $G$. Therefore, the set $\mathcal B$ of Borel subgroups of $G$ has the structure of a projective variety, where a given Borel subgroup $xBx^{-1}$ is assigned to the coset $xB$. Let $\phi_{B}$ be this mapping $\mathcal B \rightarrow G/B$ through which the isomorphism is defined.
Let $B'$ be another Borel subgroup of $G$, and write $B' = yBy^{-1}$ for some $y \in G$. I have been trying to find some isomorphism of varieties $\psi: G/B \rightarrow G/B'$ such that $$\psi \circ \phi_B= \phi_{B'}$$ which would show that the variety structure on $\mathcal B$ does not depend on the choice of initial Borel subgroup $B$. The obvious isomorphism $G/B \rightarrow G/B'$, namely the one induced from conjugation by $y$, does not satisfy $\psi \phi_B = \phi_{B'}$.
Is it the case that the variety structure on $\mathcal B$ really does depend on the choice of Borel subgroup $B$?
I am not familiar with algebraic groups or varieties, but purely on the level of abstract groups the following should take place. Let $B<G$ be a subgroup with $N_G(B)=B$. If $\mathcal{B}$ denotes the set of conjugacy classes of subgroups that $B$ is a member of, then by the orbit-stabilizer theorem we know that $\mathcal{B}\cong G/B$ as $G$-sets, where $gBg^{-1}\mapsto gB$ (which is well-defined if $B$ self-normalizes). Call this isomorphism $\phi_B$. Then it remains to see what map $\psi:G/B\to G/yBy^{-1}$ should make a commutative triangle with $\phi_B$ and $\phi_{yBy^{-1}}$.
The element $gBg^{-1}\in\mathcal{B}$ should map under $\phi_B$ to $gB$ and under $\phi_{yBy^{-1}}$ to $(gy^{-1})(yBy^{-1})$, so it would seem that $\psi$ should be conjugation-by-$y$ combined with left-multiplication-by-$y^{-1}$.