Upper triangular matrices
$
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
$
form a vector space with canonical basis:
$
e_1=\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\quad
e_2=\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
\quad
e_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}
$
that is isomorphic to $\mathbb{R}^3$ by:
$
e_1\rightarrow\vec e_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e_2\rightarrow\vec e_2=\begin{bmatrix}
0\\1\\0
\end{bmatrix}
\quad
e_3\rightarrow \vec e_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
Your linear transformation $T$ is defined as a matrix multiplication:
$
T(M)=\begin{bmatrix}
7&3\\
0&1
\end{bmatrix}
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
=
\begin{bmatrix}
7a&7b+3c\\
0&c
\end{bmatrix}
$
so, in this canonical representation, it is given by the matrix $T_e$ such that:
$
T_e\vec M=\begin{bmatrix}
7&0&0\\
0&7&3\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}=
\begin{bmatrix}
7a\\
7b+3c\\
c
\end{bmatrix}
$
Now you want a representation of the same transformation in a new basis:
$
e'_1=
\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\rightarrow\vec e'_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e'_2=\begin{bmatrix}
1&1\\
0&0
\end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix}
1\\1\\0
\end{bmatrix}
\quad
e'_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
This transformation of basis is represented by the matrices
$
S=
\begin{bmatrix}
1&1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\qquad
S^{-1}=
\begin{bmatrix}
1&-1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
$
So the matrix that represents the transformation $T$ in the new basis is:
$
T_{e'}=S^{-1}T_eS=
\begin{bmatrix}
7&0&-3\\
0&7&3\\
0&0&1
\end{bmatrix}
$
All those four cases are really just 1 case. Given vector space $U$ with basis $B = \{u_1, \dots, u_n\}$, define the coordinate isomorphism $J_B \colon \mathbb{F}^n \to U$ by
$$J_Be_j = u_j \text{ for each } j \in \{1, \dots, n\}.$$
Suppose $T \colon V \to W$ is linear, $B_1 = \{v_1, \dots, v_n\}$ is a basis of $V$, and $B_2 = \{w_1, \dots, w_m\}$ is a basis of $W$. Then the matrix representation of $T$ with respect to these bases is
$$M_{B_1}^{B_2}(T) = J_{B_2}^{-1}TJ_{B_1}.$$
So $M_{B_1}^{B_2}(T)$ takes in $B_1$-coordinates of a vector $v \in V$ and returns $B_2$-coordinates of $Tv$. Thus the $j$-th column of $M_{B_1}^{B_2}(T)$ is the $B_2$-coordinates of $Tv_j$.
Best Answer
Of course:
Note that $e_ie_j^T$ where $i,j \in \{1, \ldots, 3\}$ and $e_i$ is the standard unit basis in $\mathbb{R}^3$ is a basis for $\mathbb{R}^{3 \times 3}$.
The linear map is:
$$L\left( e_ie_j^T \right)=e_{4-i}e_{4-j}^T$$
For example:
$$L\left( e_1e_1^T \right)=e_{3}e_{3}^T$$
We can permute the entries of a matrix and it is linear, we just have to describe the image of each basis element.