= Solution
There are nontrivial proper <normal subgroups>. The <determinant> defines a surjective <group homomorphism> $G\to\{1,-1\}$. Its kernel is the <rotational symmetry group of a cube>, a normal subgroup of order twenty-four. Also $\{I,-I\}$ is a central subgroup of order two: central inversion preserves the cube and commutes with every matrix. Both provide affirmative examples.
For the body diagonal through $(1,1,1)$, write a general cube symmetry as $D_\varepsilon P$, with $P$ a <permutation matrix> and $D_\varepsilon=\operatorname{diag}(\varepsilon_1,\varepsilon_2,\varepsilon_3)$, each $\varepsilon_j=\pm1$. Since $P(1,1,1)^T=(1,1,1)^T$, preserving the unoriented diagonal requires $(\varepsilon_1,\varepsilon_2,\varepsilon_3)$ to be either $(1,1,1)$ or $(-1,-1,-1)$. Thus
$$
H=\{\varepsilon P:\varepsilon\in\{1,-1\},\ P\text{ a }3\times3\text{ permutation matrix}\}.
$$
The map $(\pi,\varepsilon)\mapsto\varepsilon P_\pi$ from $S_3\times C_2$ to $H$ is a <group homomorphism>, because central inversion commutes with every coordinate permutation. It is surjective by the displayed description and injective because a permutation matrix cannot equal the negative of another permutation matrix. Consequently the <body-diagonal stabilizer in the cube symmetry group> is
$$
\boxed{H\cong S_3\times C_2,\qquad |H|=12.}
$$
This also agrees with the <orbit-stabilizer theorem> for the transitive action on the four body diagonals. The line is preserved as a set, so exchanging its opposite endpoints is allowed.
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