Let $A\in\mathbb{R}^{n\times n}$ be diagonalisable (over $\mathbb{C}$) with pairwise distinct eigenvalues $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$, and suppose that

$$\tag{1}S^{-1}\cdot A\cdot S = \mathrm{diag}[\lambda_1,\cdots,\lambda_d]=:\Lambda \qquad \text{for some invertible } S\in\mathbb{C}^{n\times n}.$$

Assume that there is another matrix $T\in\mathbb{R}^{n\times n}$ which *'quasi-diagonalises'* $A$ in the sense that 

$$\tag{2}T^{-1}\cdot A\cdot T =\Lambda + \Delta \qquad \text{for some} \qquad \Delta\in\mathbb{R}^{n\times n} \ \text{ with } \ \|\Delta\|<\varepsilon$$
where $\varepsilon>0$ is 'small' and $\|\cdot\|$ is a matrix norm of your choice (on $\mathbb{C}^{n\times n}$).

**Question:** Can we infer that for $\varepsilon>0$ *small enough*, the matrices $S$ and $T$ are *close to one another* in the sense that     

$$\tag{3}\mathrm{inf}\{\|T\cdot S^{-1} - D\cdot P\| \mid \text{$D\in\mathbb{C}^{n\times n}$ diagonal & invertible}, \  \text{$P$ permutation}\} \ \lesssim \ \|\Delta\|$$

**In other words,** can we infer that for $\Lambda$ and $\tilde{\Lambda}:=\Lambda + \Delta$ *almost identical*, the columns of $S$ and $T$ (which define a basis for the *almost identical representations* $\Lambda$ and $\tilde{\Lambda}$ of the endomorphism $A$) *almost coincide* up to order and scale?

Any references or hints, or indeed counterexamples, are appreciated.