= Solution
Use the left <regular representation>, defined on the ordered basis $(e_I,e_R,\ldots,e_{R^5},e_{m_1},\ldots,e_{m_6})$ by $D_{\rm reg}(g)e_h=e_{gh}$. Reading the $m_4$ row of the original PDF table gives the permutation
$$
(I\ m_4)(R\ m_3)(R^2\ m_2)(R^3\ m_1)(R^4\ m_6)(R^5\ m_5).
$$
Thus an explicit $12\times12$ <permutation matrix>, in six-by-six blocks, is
$$
\boxed{D_{\rm reg}(m_4)=\begin{pmatrix}0&P\\P&0\end{pmatrix},\qquad P=\begin{pmatrix}0&0&0&1&0&0\\0&0&1&0&0&0\\0&1&0&0&0&0\\1&0&0&0&0&0\\0&0&0&0&0&1\\0&0&0&0&1&0\end{pmatrix}.}
$$
Each column has its one in the row labelled by the left product; this specifies the convention completely. Since $P^2=I_6$, the full matrix squares to $I_{12}$ and is its own inverse. Therefore for every integer, including negative integers,
$$
\boxed{[D_{\rm reg}(m_4)]^n=\begin{cases}I_{12},&n\text{ even},\\D_{\rm reg}(m_4),&n\text{ odd}.\end{cases}}
$$
This is also immediate from $m_4^2=I$ and the multiplication law for a representation.
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