Expressing a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ in terms of a new base In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ appears as
$$\phi{(x,y,z,t)} = (t,x,y,z)\ .$$ Now we are given a new base $$R = \{ (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) \}$$ of $\mathbb{R}^4$ and should find the matrix $M_R(\phi)$.  
Is my thinking correct? As I understand  I need to find the matrix created by the linear transormation: 
$$
    \left( \begin{matrix}
    0 & 0 & 0 & 1\\
    1 & 0 & 0  & 0\\
    0& 1& 0  & 0\\
0& 0& 1  & 0\\
    \end{matrix} \right)
$$
then, solve 
$$ \left[
\begin{array}{cccc|c}
  1&0&0&0&a\\
   0&1&1&1&b\\
 1&0&1&1&c\\
 0&1&0&1&d\\
\end{array}
\right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
 A: The most straight-forward way (as I see it) would be to follow this:

The columns of the matrix representation of a linear transformation are the images of the basis vectors.

So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(\phi)$ is $\phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
A: The matrix $P = \begin{bmatrix}
 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 \\
 0 & 1 & 0 & 1 \\
\end{bmatrix}$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(\phi) = P^{-1}M_E(\phi) P = \begin{bmatrix}
 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 \\
 0 & 1 & 0 & 1 \\
\end{bmatrix}^{-1}
\begin{bmatrix}
 0 & 0 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
\end{bmatrix}\begin{bmatrix}
 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 \\
 0 & 1 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
 0 & 1 & 0 & 1 \\
 1 & 0 & -1 & 0 \\
 0 & 0 & -1 & -1 \\
 0 & 0 & 2 & 1 \\
\end{bmatrix}$$
