Let G is a linear algebraic group over algebraic closed field, B is an Borel subgroup of G. Does there exist g$\in$G which is only in a finite numbers of conjugates of B (they are also Borel subgroups) ?
I choose this version of condition from the book:Tauvel, Patrice, and W. T. Rupert. Lie algebras and algebraic groups 28.2.1.
It appears in the lemma before the density theorem of Borel subgroups, but I do not see any book do this for Borel subgroups directly, they all do it for Cartan subgroups, then use Borel subgroups to cover it.
I concur with Nicolas Hemelsoet. Let's assume that $G$ is a reductive group over $\overline{k}$. Let's fix some maximal torus $T$ of $G$. Choose a regular element $t\in T(\overline{k})$. Note that if $B$ is a Borel of $G$ such that $t\in B(\overline{k})$ then writing $B=T'\rtimes U$ for $T$ a maximal torus of $G$ and $U=R_u(B)$ we evidently see that $t\in T'$. Since $t$ is regular in $T$ this implies that $T=T'$. Thus, $B$ is a standard Borel relative to $T$. We know that the choice of such objects amounts to the choice of a set of positive/simple roots in the root system of $G$ relative to $T$, and we know that such data is acted on simply transitively by $W(G,T)$. So, the number of $B$ containing $t$ is $|W(G,T)|$.