Is a "central" extension of $\mathbb{Z}/m\mathbb{Z}$ by $\mathrm{GL}_n$ necessarily split? Let $m \ge 1$ be an integer, let $k$ be a field of characteristic $0$, and let
$$
1 \rightarrow \mathrm{GL}_n \rightarrow E \rightarrow \mathbb{Z}/m\mathbb{Z} \rightarrow 1
$$
be an extension of $k$-group schemes. Since $E$ acts on $\mathrm{GL}_n$ by conjugation, there is an induced $k$-group scheme homomorphism $\mathbb{Z}/m\mathbb{Z} \rightarrow \mathrm{Out}(\mathrm{GL}_n) = \mathrm{Aut}(\mathrm{GL}_n)/\mathrm{Inn}(\mathrm{GL}_n)$. Suppose that this homomorphism is trivial, i.e., that the conjugation action of $E$ on $\mathrm{GL}_n$ is by inner automorphisms. Does this imply that the extension is split, i.e., that $E \cong \mathrm{GL}_n \times \mathbb{Z}/m\mathbb{Z}$ compatibly with the extension structure?
 A: To make my comment more precise, suppose that $k$ contains no primitive $2m$-th roots of $1$, where $m=2^j$ for some $j \ge 1$. Then we can form the central product of $G={\rm GL}_n(k)$ with a cyclic group $\langle x \rangle$ of order $2^{j+1}$, where we amalgamate $-I_n$ with the element of order $2$ in the cyclic group.
Formally, this is the quotient $E = (G \times \langle x \rangle)/\langle (-I_n,x^{2^j})\rangle$ of the direct product by a central subgroup of order $2$. With the natural embedding $G \hookrightarrow E$, we have $E/G \cong {\mathbb Z}/m{\mathbb Z}$, but we do not have $E \cong G \times {\mathbb Z}/m{\mathbb Z}$, because there is no suitable element of order $m$ centralizing $G$.
In general, with your hypothesis, the centralizer of $G$ in $E$ is equal to $\langle Z(G),x \rangle$, where $Z(G) \cong k^\times$ consists of the scalar matrices, and $x^m \in Z(G)$. So  the full extension splits as a direct product depends if and only if the extension $1 \to Z(G) \to C_E(G) \to {\mathbb Z}/m{\mathbb Z}$ splits, and this depends on $k^\times$. In particular, if $k$ is algebraically closed or, more generally, if all nonzero elements of $k$ have $m$-th roots in $k$, then all such extensions split as direct prodcuts. 
A: In the category of commutative group schemes over a field $k$, $Ext^1(\mathbb{Z}/m\mathbb{Z},\mathbb{G}_m)=k^\times/k^{\times m}$ (write the $Ext$ sequence for 
$0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/m\mathbb{Z}\to 0$). This takes care of the  case $n=1$, and the argument in Holt's answer suggests that the general case is essentially the same.
