I found the following transformation matrix for the transpose of a $2 \times 2$ matrix in $\mathbb{R}^{2 \times 2}$ (vector space of the $2 \times 2$ matrices with real numbers as elements):
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
The problem is, this is the transformation matrix with respect to a new base where the elements of the $2 \times 2$ matrix are represented as a vector in $\mathbb{R}^4$(vector space of the $4 \times 1$ matrices (vector with 4 inputs)).
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In the question, there is asked for a transformation matrix with respect to the base $\mathbb{R}^{2 \times 2}$. But I don't think there is such a matrix in $\mathbb{R}^{2 \times 2}$? So to clarify:
I have the transformation matrix with respect to the vectorspace $\mathbb{R}^4$. I now want the transformation matrix with respect to $\mathbb{R}^{2 \times 2}$. -
The following question is to find the eigenvalues of the matrix, not by calculation but with thinking. I think it has something to do with this.
Best Answer
There is an obvious correspondence between $\mathbb R^{2 \times 2}$ and $\mathbb R^4$. A matrix, by design, has each row and column corresponding to (not identical to) a basis vector in the domain or range vector space.
Therefore, your matrix is appropriate for $\mathbb R^{2 \times 2}$. You just need to keep in mind what the first basis vector is, the second, and so on.
If your actual transformation matrix is: $$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$
and we take the obvious basis for $\mathbb R^{2 \times 2}$:
$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$
$$\begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix}$$
$$\begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
then the $4 \times 4$ transformation matrix means that the first $2 \times 2$ basis matrix goes to itself, the second and the third go to each other, and the last goes to itself. The same is true for any multiples of these matrices. In other words, that $4 \times 4$ transformation matrix really does represent the transpose of a $2 \times 2$ matrix.
ADDED AFTER COMMENTS:
There is no $2\times 2$ matrix $A$ for which $A \cdot B=B^T$ for all $B$. If there were, $B^T\cdot B^{-1}$ would be it, but that expression is not constant for all $B$. So, although transposition of $2\times 2$ matrices is a linear transformation, it is represented by a $4\times 4$ matrix, not a $2\times 2$ matrix.
The eigenvectors of the transposition transformation can indeed by found by thinking. For example, you probably already know about symmetric matrices, where $A^T=A$. This means $1$ is an eigenvalue, so which matrices are symmetric? We easily get these symmetric matrices:
$$\begin{bmatrix} a & b \\ b & d \\ \end{bmatrix}$$
That clearly is a subspace of dimension $3$, and you should easily figure out a basis.
You probably also already know about skew-symmetric matrices, where $A^T=-A$. This means that $-1$ is also an eigenvalue of the transposition transformation, and we find that the skew symmetric matrices are
$$\begin{bmatrix} 0 & -c \\ c & 0 \\ \end{bmatrix}$$
This is a one-dimensional subspace, with an obvious basis.
We found a $3$-dimensional subspace and a $1$-dimensional one. There is no room for any others for our $4\times 4$ transformation, so we have found all the eigenvalues.