$T:\Bbb R^2 \to \Bbb R^2$, linear, diagonal with respect to any basis. Is there a linear transformation from $\Bbb R^2$ to $\Bbb R^2$ which is represented by a diagonal matrix when written with respect to any fixed basis?
If such linear transformation $T$ exists, then its eigenvector should be the identity matrix for any fixed basis $\beta$ of $\Bbb R^2$.
Then, I don't see, if this is possible or not.

If the transformation $T$ is represented by the matrix $A$ in basis $\mathcal{A}$, then it is represented by the matrix $PAP^{-1}$ in basis $\mathcal{B}$, where $P$ is the invertible change-of-basis matrix.
Suppose that $T$ is represented by a diagonal matrix in any basis.  Let $P$ be an arbitrary invertible matrix and $A$ any diagonal matrix:
$$P = \left[\begin{array}{cc} p_{1,1} & p_{1,2} \\ p_{2,1} & p_{2,2} \end{array}\right] \text{ and } A = \left[\begin{array}{cc} d_1 & 0 \\ 0 & d_2 \end{array}\right].$$
Now, calculate
$PAP^{-1} = \dfrac{1}{\det P} \left[\begin{array}{cc} b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \end{array}\right]$, where the entries $b_{i,j}$ are polynomials in the $p_{i,j}$ and $d_i$ variables.
For this new conjugated matrix to be diagonal, we have the following two equations.  (Check!)
$$\begin{align*}
0 = b_{1,2} &= (d_2 - d_1)p_{1,1}p_{1,2} \\
0 = b_{2,1} &= (d_1 - d_2)p_{2,1}p_{2,2}
\end{align*}$$
Since $P$ is arbitrary, the only way for these equations to always be satisfied is for $d_1 = d_2$.  In other words, the original matrix $A$ was a scalar multiple of the identity.
$$A = d \cdot \operatorname{Id}_2 = \left[\begin{array}{cc} d & 0 \\ 0 & d \end{array}\right].$$
