Let $T$ be the usual maximal torus of $G = \textrm{GL}_n$. The center $Z_G$ of $\textrm{GL}_n$ is connected, so if we look at the exact sequence
$$1 \rightarrow Z_G \rightarrow T \rightarrow T/Z_G \rightarrow 1 ,$$
the corresponding exact sequence of abelian groups
$$0 \rightarrow X(T/Z_G) \rightarrow X(T) \rightarrow X(Z_G) \rightarrow 0$$
splits, because $X(Z_G) \cong \mathbb{Z}$ is free. The equivalence of categories between tori and finitely generated free abelian groups tells us that our original exact sequence splits as well.
What I want to do is get a "nice" splitting, one which can be found by identifying $T$ with the almost direct product $(T_d \times Z_G)/K$, where $T_d$ is the usual maximal torus of $\textrm{SL}_n$, and $K = T_d \cap Z_G$ is the subgroup generated by $(\zeta_n I_n, \zeta_n I_n)$, where $\zeta_n$ generates the group of $n$th roots of unity. Here, we can use the second isomorphism theorem to identify $T/Z_G = T_d/ \langle \zeta_n I_n \rangle$.
So, what I want is to be able to discern a nice splitting from the exact sequence
$$1 \rightarrow Z_G \rightarrow (T_d \times Z_G)/K \rightarrow T_d/\langle \zeta_n I_n \rangle$$
This is because later, I want to replace $Z_G$ with a larger torus $\widehat{Z_G}$, and embed $\textrm{GL}_n$ into a larger reductive group $(G \times \widehat{Z_G})/K$ with center $\widehat{Z_G}$ and maximal torus $(T_d \times \widehat{Z_G})/K$, and do a similar procedure.
I have begun the special case $n = 4$:
Assume we are not in characteristic two. Every $t \in T_d$ can be written uniquely as $\textrm{diag}(t_1, t_2t_1^{-1}, t_3t_2^{-1}, t_3^{-1})$. Let $f_i : i = 1, 2, 3$ be the basis for $X(T_d)$ sending $t$ to $t_i$.
Let $N = \langle \zeta_4 I \rangle = \langle i I \rangle$. We can identify $X(T_d/N) = X(T/Z_G)$ with those elements of $X(T_d)$ whose kernel contains $iI$. So we see that $4f_1, 2f_2, 4f_3$ is a basis of $X(T_d/N)$.
Similarly, $X(T_d \times Z_G/K)$ consists of those characters of $X(T_d \times Z_G) = X(T_d) \oplus X(Z_G)$ whose kernel contains $(iI,iI)$. Let $f$ be the usual generator of $X(Z_G)$. Then for $\chi = c_1f_1 + c_2f_2 + c_3 f_3 + cf$ in $X(T_d \times Z_G/K)$, we have
$$1 = \chi(iI,iI) = i^{c_1 + 2c_2 + 3c_3 + c}$$
and so $c_1 + 2c_2 + 3c_3 + c \equiv 0 \pmod{4}$.