= Solution
Since $P=P'=0$, it suffices to map the two edge <vectors>. Let the columns of the required <matrix> be $u_1,u_2,u_3$. The condition $MQ=Q'$ gives
$$
u_2=\left(-\frac12,0,\frac{\sqrt3}{2}\right)^T.
$$
The condition $MR=R'$ then gives
$$
\frac{\sqrt3}{2}u_1+\frac12u_2=R',
$$
and hence
$$
u_1=\left(\frac{\sqrt3}{2},0,\frac12\right)^T.
$$
These are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0)^T$ produces the orthogonal <matrix>
$$
\boxed{
M=
\begin{pmatrix}
\sqrt3/2&-1/2&0\\
0&0&-1\\
1/2&\sqrt3/2&0
\end{pmatrix}}.
$$
Its columns are orthonormal, and direct substitution verifies $MP=P'$, $MQ=Q'$, and $MR=R'$.
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