= Solution
Use <change of basis> matrices whose columns are the respective <basis vectors> in standard coordinates:
$$
S_B=\begin{pmatrix}1&1\\1&-1\end{pmatrix},\quad
S_C=\begin{pmatrix}1&0&0\\1&1&1\\0&0&1\end{pmatrix},\quad
S_{B'}=\begin{pmatrix}0&2\\2&0\end{pmatrix},\quad
S_{C'}=\begin{pmatrix}1&0&1\\0&1&2\\-1&0&1\end{pmatrix}.
$$
The coordinate relation is $[\Phi(x)]_C=A[x]_B$. Therefore the standard-coordinate <matrix of a linear map> is $T=S_CAS_B^{-1}$, and the new coordinate matrix is $A'=S_{C'}^{-1}TS_{B'}$. Multiplying gives
$$
T=\begin{pmatrix}0&1\\2&0\\0&-1\end{pmatrix},\qquad
TS_{B'}=\begin{pmatrix}2&0\\0&4\\-2&0\end{pmatrix},\qquad
\boxed{A'=\begin{pmatrix}2&0\\0&4\\0&0\end{pmatrix}.}
$$
One can also read off the columns: the two new domain <basis vectors> map to twice the first and four times the second new codomain <basis vector>. All four <change of basis> matrices are invertible. In particular the printed codomain vectors are three-dimensional; retaining their third coordinates is essential.
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