Is a basis that almost diagonalizes a matrix 'close' to its eigenbasis? 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
$$\tag{3}\mathrm{inf}\{\|T\cdot S^{-1} - D\cdot P\| \mid \text{$D\in\mathbb{C}^{n\times n}$ diagonal & invertible}, \  \text{$P$ permutation matrix}\} \ \lesssim \ \|\Delta\|$$
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.
Edit: As demonstrated in Ruy's answer, the desired conclusion doesn't apply without additional conditions on $A$; are you aware of any such conditions that are as mild as possible?
 A: As stated, the answer is no,  at least for the operator norm.
For a counter example choose any $\varepsilon >0$,
and
take $A$ to be a
diagonal matrix whose diagonal entries $\lambda _1, \lambda _2, \ldots , \lambda _n$ are all distinct,  but satisfy
$$
  |\lambda _i-1|<\varepsilon/2 ,\quad \forall i=1,\ldots ,n.
  $$
Since $A$ is already diagonal, we can take $S=I$  and $\Lambda =A$, in order to satisfy the required relation:
$$
  S^{-1}AS=\Lambda .
  $$
Choose any unitary operator $T$,
whatsoever, and notice that
$$
  \|T^{-1}AT - \Lambda \| =
  \|T^{-1}AT - A\| \leq  $$$$\leq
  \|T^{-1}AT - I\| +  \|I - A\| =
  \|T^{-1}(A-I)T \|  +  \|I - A\| = $$$$ =
  2\|A-I\| < \varepsilon,
  $$
so all of the required conditions are verified.
Neverteless,
given the arbitrary choice of $T$, there is certainly no reason to expect  the existence of a diagonal matrix $D$,
and a permutation matrix $P$, such that
$$
  \|TS^{-1} - D P\| =   \|T - D P\|
  $$
is small.
I suspect that one needs to require the eigenvalues of $A$ to be spread far apart if one is to hope for an affirmative
answer.
