= Solution
Identify $\mathfrak t$ with the coordinate vector $t=(t_1,\ldots,t_n)$ through $H(t)$. A <root reflection> is $s_\alpha(H)=H-\alpha(H)\alpha^\vee$. Since $(\varepsilon_i\pm\varepsilon_j)^\vee=H_i\pm H_j$, for $i<j$ the two possibilities are
$$
\boxed{\begin{aligned}s_{\varepsilon_i-\varepsilon_j}(t)_i&=t_j,&s_{\varepsilon_i-\varepsilon_j}(t)_j&=t_i,\\s_{\varepsilon_i+\varepsilon_j}(t)_i&=-t_j,&s_{\varepsilon_i+\varepsilon_j}(t)_j&=-t_i.
\end{aligned}}
$$
Every other coordinate is unchanged, and $s_{-\alpha}=s_\alpha$, so these formulas cover every <root>. The complete diagonal matrix is reconstructed with its last $n$ entries the reversed negatives of the first $n$ entries.
The <Weyl group of Dn> consists of all permutations of the $n$ coordinates together with an <even number> of coordinate sign changes. Equivalently,
$$
\boxed{W(D_n)=\{t\mapsto(\delta_i t_{\sigma(i)})_{i=1}^n:\sigma\in S_n,\ \delta_i\in\{\pm1\},\ \prod_i\delta_i=1\}\cong(\mathbb Z/2\mathbb Z)^{n-1}\rtimes S_n.}
$$
Its order is $2^{n-1}n!$. The difference-root reflections give ordinary transpositions; composing a sum-root reflection with the corresponding difference-root reflection negates precisely the two selected coordinates. These describe the generators of all even signed permutations. For $n=2$ the group has four elements; for the toral case $n=1$ it is trivial.
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