Is there a matrix that can be used to find the transpose of a matrix? Let $A$ be a general $n\times n$ invertible matrix. Let $T^A$ be the "transposer" matrix i.e. $T^A A = A'$. (Does that $T^A$ multiplied by $A$ equal the transpose of $A$?) Then does $T^A$ depend on the matrix $A$: is $T^A = T^B$ for all invertible Matrices? Prove your claim.
So can you take a matrix times the given matrix in order to find its transpose? If so, is that a general form that can be used for all invertible matrices? 
I'm guessing it does not, but I am not totally sure or know how to go about proving that and I cannot find anything online about it.
 A: There is no general transposer matrix.. For this, e.g. note that the only transposer for the identity in any dimension is the identity as the equation
$$T^II=I$$
needs to be fulfilled, but $T^II=T^I$.
However, for e.g. for non-symmetric matrices $A$, the identity is definitely not the transposer, as $A^\top\neq A$, but $IA=A$.
Now, we may try to construct the transposer matrix by considering $X=(x_{ij})_{i,j\leq n}$ and $A=(a_{ij})_{i,j\leq n}$, and then solving the equation $XA=A^\top$ for $X$. I.e. by the laws of matrix multiplication, you have to have
$$a_{ij}=\sum_{k=1}^n a_{jk}x_{ik}$$
for all $i,j$. However, this system of linear equations is not always solvable (only if $A$ is not symmetric). For this, consider the following example:
Look at $A=\begin{pmatrix}0 &1\\0 &0\end{pmatrix}$, then
$$\begin{pmatrix}x &y\\z &w\end{pmatrix}\begin{pmatrix}0 &1\\0 &0\end{pmatrix}=\begin{pmatrix}0 &x\\0 &z\end{pmatrix}\neq\begin{pmatrix}0 &0\\1 &0\end{pmatrix}$$
A: Even in the simple case of $2\times 2$ matrices such a general transposer matrix does not exist. 
Note that in order to keep the diagonal terms unchanged you need the so called transposer to be the  identity matrix which in general does not change the matrix to its transpose.
A: Let $A$ be an $n \times n$ matrix.  For any such matrix, you can construct an $n \times n$ transposer matrix $T$ so that $TAT = A^T$.
The Singular Value Decomposition tells us that we can find two orthonormal bases for $\mathbb{R}^n$: $v_1, v_2.. v_n$ and $u_1, u_2,.. u_n$ so that $A v_i = \sigma_i u_i$.  In matrix form $A = U \Sigma V^T$ where $U$ and $V$ have the $u's$ and $v's$ as their columns and $\Sigma$ is a diagonal matrix of the singular values $\sigma_i$.
Then $A^T = V \Sigma U^T$.  Since $U$ and $V$ are matrices of orthonormal columns, they are their own inverses, so:
$A^T = V \Sigma U^T = V (U^T U) \Sigma (V^T V) U^T = V U^T (U \Sigma V^T ) V U^T = (V U^T) A (V U^T)$
So the originally claimed transposer matrix $T = VU^T$
A: Let's test the $2$-dimensional case.
Let $A=\left[\begin{matrix}a&b\\
c &d\end{matrix}\right]$.
Then any such $T^A=\left[\begin{matrix}t_{11}&t_{12}\\
t_{21} &t_{22}\end{matrix}\right]$ must obey:
$$t_{11}a+t_{12}c=a$$
$$t_{11}b+t_{12}d=c$$
$$t_{21}a+t_{22}c=b$$
$$t_{21}b+t_{22}d=d$$
Solve this system for the $t_{ij}$ and you will obtain a result based on the entries in $A$, answering your question.
