Let $X$ be a smooth quasiprojective variety over $\mathbb{C}$. It has a group of (algebraic) automorphisms $
\DeclareMathOperator{\Aut}{Aut}
\Aut(X)$.

Define a _torus in $\Aut(X)$_ to be a faithful algebraic action $\alpha: (\mathbb{C}^*)^m \times X \to X$ of $T := (\mathbb{C}^*)^m$. It is _maximal_ if there is no torus $T' \cong (\mathbb{C}^*)^{m+1}$ such that the action can be faithfully extended to $T'$.

Suppose $T_1$, $T_2$ are two maximal tori. Are they necessarily conjugate, i.e. does there exist a $\gamma \in \Aut(X)$ and an isomorphism of algebraic groups $\rho: T_1 \to T_2$ such that $\gamma \circ \alpha_1 \circ \gamma^{-1} = \alpha_2(\rho(-), -)$ ?

Most people seem to believe this is false. Are there any known counterexamples?

**Remark:** The rank of a torus is bounded by $(\dim X)^2 + 2 \dim X$. As a consequence, every torus in $\Aut(X)$ is contained in a maximal torus.  
_Proof:_  An algebraic action of $(\mathbb{C}^*)^m$ gives us a smooth action of the compact Lie group $\mathbb{T}^m := (S^1)^m$. Picking a Riemannian metric $g$ on the smooth manifold $X$ and averaging by $\mathbb{T}^m$ gives us a metric $\bar{g}$ which is $\mathbb{T}^m$-invariant, so 
$
\DeclareMathOperator{\Isom}{Isom}
\mathbb{T}^m \subset \Isom(X, \bar{g})
$.
But isometries are determined by their 1-jets so $m \leq \dim \Isom(X, \bar{g}) \le (2 \dim X)^2 + 2 \dim X$. If we further take into account that the isometries in question are holomorphic we can improve the bound to $(\dim X)^2 + 2 \dim X$.