Consider a diagonalizable linear operator $T$, and let $\{\alpha\}$ be the basis of eigenvectors of $T$. Let $\{\beta\}$ be the standard basis for $\mathbb R^n$ and let $\{\gamma\}$ be another arbitrary basis. Now the similarity transformation when the $T$ is written in standard orders basis proceeds as follows -
\begin{equation} S^{-1}[T]_\beta S = [S]_\beta^\alpha [T]_\beta^\beta [S]_\alpha^\beta = [STS]_\alpha^\alpha........................(1) \end{equation}
Equation $(1)$ leads to a diagonal representation of $T$.
Now consider the similarity transformation when $T$ is written in the basis $\{\gamma\}$. This proceeds as follows :
\begin{equation} S’^{-1}[T]_\gamma S’= [S’]_\gamma^\alpha [T]_\gamma^\gamma [S’]_\alpha^\gamma = [S’TS’]_\alpha^\alpha........................(2) \end{equation}
Equation $(2)$ doesn’t lead to a diagonal matrix.
The question is simple - why ?
For e.g.:
Let $T(x,y) = (2x+y,11x+12y)$. In the standard basis this is $T = \begin{pmatrix}2 &1 \\ 11 & 12\end{pmatrix}$. The basis of eigenvectors is $[1,11],[-1,1]$ in which the matrix is $\begin{pmatrix} 13 &0 \\ 0 & 1 \end{pmatrix}$. Let $B$ be a new basis, $\{[1,1],[1,0]\}$. Then in this basis one sees the matrix for $T$ is $M=\begin{pmatrix} 23 & 11 \\ -20 & -9 \end{pmatrix}$. Now, the change of basis from $B$ to the eigenbasis is the matrix $S = \begin{pmatrix} 1 & -1 \\ 11 & 1\end{pmatrix}^{(-1)} \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix} = \frac 1{12} \begin{pmatrix} 2 & 1\\ -10 &-11\end{pmatrix}$. Now, $S^{-1} MS$ should be diagonal but is not : both the off-diagonal entries is non-zero.
Why does the Similarity Transformation not lead to a diagonal matrix when the the linear operator is written in a basis other than the standard ordered basis?
Why is the standard ordered basis special in getting a diagonal matrix using a similarity transformation?
The symbols you use don't make sense and they are very confusing.
When $\alpha$ and $\beta$ are bases of two finite-dimensional vector spaces $V$ and $W$ respectively and $T:V\to W$ is a linear map, the matrix representation of $T$ with respect to the bases $\alpha$ and $\beta$ is denoted by $[T]_\alpha^\beta$. Note that $T$ is a linear map, not a matrix, and $[T]_\alpha^\beta$ is not a linear map, but a matrix representing a linear map.
When one writes expressions like $[S]_\alpha^\beta,[STS]_\alpha^\alpha$ or $S^{-1}[T]_\beta S$, unless one overloads the same symbol to mean different things, $S$ should mean either a linear map or a matrix. However, if $S$ is a matrix, it doesn't make any sense to write $[S]_\alpha^\beta$ or $[STS]_\alpha^\alpha$. Similarly, if $S$ is a linear map, it doesn't make sense to write $S^{-1}[T]_\beta S$. The symbol $S$ in your equality $(1)$ thus have contradictory meanings.
A correct and clearer way to deal with matrix representations under changes of bases is to stick with the convention that the $T$ in $[T]_\alpha^\beta$ is a linear map. When $\alpha$ and $\beta$ are bases of the same vector space $V$, the change of basis matrix from $\alpha$ to $\beta$ is $[\operatorname{id}]_\alpha^\beta$ --- since we only change the basis of $V$ without mapping any vector in $V$ to a different vector, the linear map inside the pair of square brackets is the identity map $\operatorname{id}$, not some mysterious $S$. Given the matrix representation $[T]_\alpha^\alpha$ of a linear operator $T$ with respect to the basis $\alpha$, the matrix of $T$ with respect to another basis $\beta$ is obtained by $$ [T]_\beta^\beta =[\operatorname{id}\circ T\circ \operatorname{id}]_\beta^\beta =[\operatorname{id}]_\alpha^\beta [T]_\alpha^\alpha [\operatorname{id}]_\beta^\alpha =\left([\operatorname{id}]_\beta^\alpha\right)^{-1} [T]_\alpha^\alpha [\operatorname{id}]_\beta^\alpha. $$
When $\beta$ is an eigenbasis for $T$, the matrix $[T]_\beta^\beta$ is diagonal. Hence $[T]_\alpha^\alpha=[\operatorname{id}]_\beta^\alpha[T]_\beta^\beta \left([\operatorname{id}]_\beta^\alpha\right)^{-1}$ is a diagonalisation of $[T]_\alpha^\alpha$. This is true regardless of whether $\alpha$ is the standard basis or not.
In your example, let $\alpha$ be the standard basis, $\beta$ be the eigenbasis from $T$ and $\gamma$ be the new basis. Then $$ [\operatorname{id}]_\beta^\alpha=\pmatrix{1&-1\\ 11&1}, [\operatorname{id}]_\gamma^\alpha=\pmatrix{1&1\\ 1&0}. $$ Hence the matrix representation of $T$ with respect to $\gamma$ is $$ [T]_\gamma^\gamma =\left([\operatorname{id}]_\gamma^\alpha\right)^{-1} [T]_\alpha^\alpha [\operatorname{id}]_\gamma^\alpha =\pmatrix{1&1\\ 1&0}^{-1}\pmatrix{2&1\\ 11&12}\pmatrix{1&1\\ 1&0} =\pmatrix{23&11\\ -20&-9}. $$ The change-of-basis matrices from $\gamma$ to $\beta$ and from $\beta$ to $\gamma$ are given by \begin{align} [\operatorname{id}]_\gamma^\beta &=[\operatorname{id}]_\alpha^\beta[\operatorname{id}]_\gamma^\alpha =\left([\operatorname{id}]_\beta^\alpha\right)^{-1}[\operatorname{id}]_\gamma^\alpha =\pmatrix{1&-1\\ 11&1}^{-1}\pmatrix{1&1\\ 1&0} =\frac{1}{12}\pmatrix{2&1\\ -10&-11},\\ [\operatorname{id}]_\beta^\gamma &=\left([\operatorname{id}]_\gamma^\beta\right)^{-1} =\pmatrix{11&1\\ -10&-2}. \end{align} One can verify that the similarity transform $$ [\operatorname{id}]_\gamma^\beta [T]_\gamma^\gamma [\operatorname{id}]_\beta^\gamma =\frac{1}{12}\pmatrix{2&1\\ -10&-11}\pmatrix{23&11\\ -10&-9}\pmatrix{11&1\\ -10&-2}=\pmatrix{13&0\\ 0&1} $$ indeed a diagonalises $[T]_\gamma^\gamma$.