Permute columns by pre-multiplying and rows by post-multiplying? I was looking at Gilbert Strang's lectures on Linear Algebra and noticed that in lecture 2, Elimination with Matrices, around the 40nth minute he mentions that you can use the permutation matrix, 
$$P=
\begin{bmatrix}
    0       & 1 \\
    1       & 0 
\end{bmatrix}
$$
so that $AP$ is a permutation of $A$'s columns and $PA$ is a permutation of $A$'s rows.
I was wondering if there exists a matrix $P'$ such that $P'A$ is a permutation of $A$'s columns and $AP'$ is a permutation of $A$'s rows.


*

*How to prove there is no $P'$ such that $P'A=AP$ and $AP'=PA$ for all
$n\times n$ matrices?

*For which matrices $A$ there is such a $P'$?



Tried the $2\times 2$ case
$L$ pre-multiplies $A$ and permutes its columns,
$$
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4 
\end{bmatrix}
\begin{bmatrix}
a       & b \\
c       & d 
\end{bmatrix}=\begin{bmatrix}
b       & a \\
d       & c 
\end{bmatrix} \Rightarrow
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4 
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
bd-ac       & a^2-b^2 \\
d^2-c^2       & ac-bd
\end{bmatrix}=L
$$
$R$ post-multiplies $A$ and permutes its rows,
$$
\begin{bmatrix}
a       & b \\
c       & d 
\end{bmatrix}
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4 
\end{bmatrix}=
\begin{bmatrix}
c       & d \\
a       & b 
\end{bmatrix} \Rightarrow
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4 
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
cd-ab       & d^2-b^2 \\
a^2-c^2       & ab-cd
\end{bmatrix}=R
$$


*

*$L\neq R$ in the general case

*$det(A)\neq 0$ for $R$ and $L$ to exist 

*if $det(A)\neq 0$ and $a=d=0$, then $P'=L=R$

Additional notes
I feel I have to clarify further. I was looking for a matrix $P'$ that behaves just like $P$ but from the opposite side, meaning $P'$ permutes columns when pre-multiplying $A$ and rows when post-multiplying $A$.

$\newcommand\bm\boldsymbol$
If such $\bm P'$ works for all matrices, then for all $\bm A$, 
$$
\bm {AP} = \bm {P'A}, 
$$ 
then specifically it works for the identity matrix $\bm I$, i.e.
$$
\bm {IP} = \bm {P'I}, 
$$
then the only candidate of $\bm P'$ is $\bm P$ again. But clearly 
$$
\bm {PA} = \bm {AP} 
$$
only holds for specific matrices $\bm A$. 
Conclusion: maybe for some $\bm A$, there exists $\bm P'$ that $\bm {P'A}$ swap two columns of $\bm A$, but there exists no universal matrices $\bm P'$. 
