Is there a linear transformation that would flip the elements of a matrix around its center? Is there a linear transformation that would move the elements of a matrix in the following fashion? or some combination of other matrix operations including the transpose etc?
$$ 
\begin{gather}
 \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}  \end{bmatrix}
 \xrightarrow{\;\;T\;\;}
 \begin{bmatrix} \color{red}{a_{33}} & \color{blue}{a_{32}} & \color{green}{a_{31}} \\ \color{orange}{a_{23}} & a_{22} & \color{orange}{a_{21}} \\ \color{green}{a_{13}} & \color{blue}{a_{12}} & \color{red}{a_{11}}  \end{bmatrix}
\end{gather}
$$
 A: 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.
A: Fix a field $\mathbb{K}$ and $n\in\mathbb{Z}_{>0}$.  Then, $[n]:=\{1,2,\ldots,n\}$ and $\text{Mat}_{n\times n}(\mathbb{K})$ is the vector space of $n$-by-$n$ matrices over $\mathbb{K}$.  Let $T:\text{Mat}_{n\times n}(\mathbb{K})\to\text{Mat}_{n\times n}(\mathbb{K})$ be the flipping map sending  an element $\big[a_{i,j}\big]_{i\in[n],j\in[n]}$ of $\text{Mat}_{n\times n}(\mathbb{K})$ to $\big[a_{n+1-i,n+1-j}\big]_{i\in[n],j\in[n]}$.  It is actually clear that $T$ is a $\mathbb{K}$-linear map.  However, I believe that the OP wants more.
If $P$ denotes the $n$-by-$n$ permutation matrix
$$\newcommand\iddots{\mathinner{
  \kern1mu\raise1pt{.}
  \kern2mu\raise4pt{.}
  \kern2mu\raise7pt{\Rule{0pt}{7pt}{0pt}.}
  \kern1mu
}}\begin{bmatrix}0&0&\dots&0&0&1\\
0&0&\dots&0&1&0\\
0&0&\dots&1&0&0\\
\vdots&\vdots&\iddots&\vdots&\vdots&\vdots\\0&1&\dots&0&0&0\\
1&0&\dots&0&0&0\end{bmatrix}\,,$$
then $T$ is the same as conjugation by $P$.  That is, $$T(A)=PAP^{-1}$$ for all $A\in\text{Mat}_{n\times n}(\mathbb{K})$.  (Since $P^{-1}=P$, we can also write $T(A)=PAP$ for all $A\in\text{Mat}_{n\times n}(\mathbb{K})$.)  In particular, when $n:=3$, we have
$$
 \begin{bmatrix}  0& 0 & 1 \\ 0 &1 & 0 \\ 1 & 0 & 0  \end{bmatrix}\,
\begin{bmatrix} \color{red}{a_{11}} & \color{blue}{a_{12}} & \color{green}{a_{13}} \\ \color{orange}{a_{21}} & \color{grey}{a_{22}}  & \color{magenta}{a_{23}} \\ \color{teal}{a_{31}} & \color{cyan}{a_{32}} & \color{brown}{a_{33}}  \end{bmatrix}\,
 \begin{bmatrix}  0& 0 & 1 \\ 0 &1 & 0 \\ 1 & 0 & 0  \end{bmatrix}
=
 \begin{bmatrix} \color{brown}{a_{33}} & \color{cyan}{a_{32}} & \color{teal}{a_{31}} \\ \color{magenta}{a_{23}} & \color{grey}{a_{22}} & \color{orange}{a_{21}} \\ \color{green}{a_{13}} & \color{blue}{a_{12}} & \color{red}{a_{11}}  \end{bmatrix}\,.
$$
