Subgroups of $SL(2,\mathbb{Z})$. Is every finite index subgroup $\Gamma$ of $\text{SL}_2(\mathbb{Z})$ determined up to conjugacy by: $$(d,e_4,e_6,e_{-1},c_1,\ldots,c_r),$$ where $d$ is the index of $\Gamma$, $e_4,e_6$ its number of conjugacy classes of elements of order 4 and 6, $e_{-1}$ is 0 or 1 according to if $-I\in\Gamma$, and $c_1,\ldots,c_r$ are its cusp widths. 
I guess $d$ is redundant since $d = \sum_i c_i$.
This is basically a question about the "signature of a Fuchsian group". I remember reading somewhere that you can recover a cocompact Fuchsian group from its signature, though I don't think there was a proof. Furthermore, subgroups of $SL_2(\mathbb{Z})$ are of course not cocompact.
Does anyone have references/proofs for any results related to this?
 A: $\def\SL{\mathrm{SL}}$
$\def\Z{\mathbf{Z}}$
The answer to this question if false, but I suspect that you may have misread the original claim. You are talking about $\Gamma$ up to conjugation in $\SL_2(\Z)$, but a variant would be to ask simply about $\Gamma$ as an abstract group up to isomorphism. It is quite likely that $\Gamma$ is determined by this information, because the group theory here is much simpler (for example, if $\Gamma$ is torsion free and finite index then it is actually free, and the rank is determined by the index.) 
The basic problem for your version of the question is that $\SL_2(\Z)$ has many interesting quotients (since it is virtually free) and the data is nowhere near enough to determine these and hence distinguish the corresponding normal subgroups. There is probably a simple example one can compute by hand if computation is your sort of thing.
Let us recall that
$$\SL_2(\Z)/Z(\SL_2(\Z)) = \mathrm{PSL}_2(\Z) = \langle S, T | S^2, (ST)^3 \rangle.$$
Here
$$S  = \left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right),
\qquad
T = \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).$$
I am going to choose $\Gamma$ normal in $\SL_2(\Z)$ such that
$$G = \SL_2(\Z)/\Gamma = \langle S, T | S^2, (ST)^3, T^7,\Delta\rangle,$$
for some extra choice of relations $\Delta$. What can one say about the invariants in this case? First, let's suppose that $G$ is not the trivial group. An easy exercise shows that this forces $S$, $ST$, and $T$ to have exact orders $2$, $3$, and $7$ respectively. What can one deduce from this?
Any element in $\SL_2(\Z)$ of order $4$ is conjugate to $S$, but the image of $S$ is non-zero, so $e_4 = 0$.
Any element in $\SL_2(\Z)$ of order $6$ is conjugate to $ST$, but the image of $ST$ is non-zero, so $e_6 = 0$.
The quotient is a quotient of $\mathrm{PSL}_2(\Z)$, so $e_4 = -1$.
The image of $T$ has order $7$, and similarly the image of any conjugate of $T$ has order $7$. Thus $c_i = 7$ for all $i$.
As you noted, the number of $c_i$ is determined by the index.
Taken together, your claim would imply the following: Any finite quotient of
$$\langle S, T | S^2, (ST)^3, T^7 \rangle$$
is determined by its order.  Now finite quotients of this group are well studied;
they are known as Hurwitz groups because of the relation with automorphisms of curves of genus $\ge 2$ with maximal automorphism groups. 
It is a theorem of Higman that $A_n$ is a Hurwitz group for sufficiently large $n$.
By Goursat's Lemma, it follows that $A_n \oplus A_m$ is also a Hurwitz group for $n > m$ and $m$ sufficiently large.  But then we have the following two Hurwitz groups of the same order:
$$A_{n}, \qquad A_{n-1} \oplus A_m, \qquad n = |A_m|.$$
