do any two conjugates of [[1,1],[0,1]] generate SL(2,Z)? Do any two (distinct) conjugates of the matrix $[[1,1],[0,1]]$ generate $SL(2,\mathbb{Z})$?
Of course the conjugates $[[1,1],[0,1]]$ and $[[1,0],[1,1]]$ generate it, and some explicit computations show that a number of "randomly" chosen conjugates also generate, but I don't have a proof.
 A: It is not true --- there exist two conjugates of that matrix which do not generate $SL(2,\mathbb{Z})$.
There is an amalgamated free product presentation
$$SL(2,\mathbb{Z}) = \mathbb{Z}/4\mathbb{Z} *_{\mathbb{Z}/2\mathbb{Z}} \mathbb{Z} / 6 \mathbb{Z}
$$
Thinking of the action of $SL(2,\mathbb{Z})$ on the upper half plane by fractional linear transformations, the amalgamating subgroup $\mathbb{Z}/2\mathbb{Z}$ is the kernel of the action, the $\mathbb{Z}/4\mathbb{Z}$ subgroup is the stabilizer of $p = 1+0i$, and the $\mathbb{Z}/6\mathbb{Z}$ subgroup is the stabilizer of $q = \frac{1}{2} + \frac{\sqrt{3}}{2} i$. Letting $E = \overline{pq}$ be the hyperbolic geodesic with endpoints $p,q$, the union of the translates of $E$ under the action of $SL(2,\mathbb{Z})$ is a tree $T$ with fundamental domain $E$, the endpoint $p$ has valence~$2$ in $T$, and the endpoint $q$ has valence~$3$. This tree $T$ is the "Bass-Serre tree" of the amalgamated free product presentation.
The matrix $M = [[1,1],[0,1]]$ represents the fractional linear transformation $z \mapsto z+1$, whose restriction to the tree $T$ is a translation along a "$T$-line" $L \subset T$ that stays a uniformly bounded distance from the horocycle $y=1$. Here by a "$T$-line" I simply mean an isometrically embedded copy of the real line in $T$ with respect to the path metric on $T$.
If you choose a conjugating element $A$ which is sufficiently complicated such that the lines $A(L)$ and $L$ are disjoint in $T$ and are connected by a path consisting of at least 4 translates of the edge $E$, then $AMA^{-1}$ and $M$ generate a rank $2$ free subgroup of infinite index. I'm pretty sure the element 
$$A = [[13,8],[8,5]]
$$
is sufficiently complicated (I just took a power of $[[2,1],[1,1]])$.
Your choice of conjugating element $A$ has the property that $A(L) \cap L$ is a nonempty path consisting of two translates of $E$.
