= Solution
For each <involution> $t\in G$, its <conjugacy class> has size fifteen, so its <centralizer> $C_G(t)$ has order four. A group of order four is a <cyclic group> or a <Klein four-group>; the cyclic possibility would contain an element of order four, excluded by the geometric enumeration. Hence each $C_G(t)$ is a <Klein four-group> containing three <involutions>.
If $u$ is any of the three <involutions> in $C_G(t)$, the latter group is an <abelian group> and hence is contained in $C_G(u)$. Both <centralizers> have order four, so they are equal. It follows that the fifteen <involutions> partition into five triples, with five distinct <centralizers>. Conjugation by $G$ permutes these five subgroups and gives a <group homomorphism> $\theta:G\to S_5$. This action is nontrivial: otherwise each of these order-four subgroups would be a <normal subgroup>, contrary to simplicity. The <kernel> of $\theta$ is normal; since $G$ is simple and the action nontrivial, its <kernel> is trivial. Thus $\theta$ is injective.
The sign of $\theta(g)$ defines a homomorphism $G\to\{\pm1\}$. If it were nontrivial, simplicity would make it injective, impossible for a group of order sixty mapping into one of order two. Hence every image permutation is even, and $\theta(G)\subseteq A_5$. Both groups have order sixty, so they are equal. This constructs the requested <group isomorphism>:
$$
\boxed{\operatorname{Rot}(D)\cong A_5.}
$$
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