Distance between two images of the same element by two different linear maps Let $n \in \mathbb{N}$, and $\mathbb{M}(n)$ the set of all square matrices of size $n$, with real positive entries. We define two linear functions $T_1: \mathbb{M}(n) \to \mathbb{M}(n), A \mapsto D_1AD_1^{-1}$, and $T_2: \mathbb{M}(n) \to \mathbb{M}(n), A \mapsto D_2AD_2^{-1}$ where $D_1$ and $D_2$ are diagonal matrices. These correspond, if I am not mistaken, to a change of basis in $\mathbb{R}^n$. Now for some given $A \in \mathbb{M}(n)$, I am interested in upper bounding $|| T_1(A) - T_2(A) ||_2$ by some expression involving a control over $||D_1 - D_2||_2$ (or other measure of distance between the two diagonal matrices), and $A$ or some norm of it, as I am assuming that if $D_1$ and $D_2$ are close, and we start from the same element in $\mathbb{M}(n)$, the images cannot be too far apart either, but I am missing the tool.
Note: None of the diagonal entries of the matrices $D_1$ and $D_2$ are null for this problem.

Note that
\begin{align*}
T_{1}A-T_{2}A & =D_{1}AD_{1}^{-1}-D_{2}AD_{2}^{-1}\\
 & =D_{1}AD_{1}^{-1}-D_{2}AD_{1}^{-1}+D_{2}AD_{1}^{-1}-D_{2}AD_{2}^{-1}\\
 & =\left(D_{1}-D_{2}\right)AD_{1}^{-1}+D_{2}A\left(D_{1}^{-1}-D_{2}^{-1}\right).
\end{align*}
Therefore,
$$
\left\Vert T_{1}A-T_{2}A\right\Vert \leq\left(\Vert D_{1}-D_{2}\Vert \Vert D_{1}^{-1}\Vert +\Vert D_{2}\Vert \Vert D_{1}^{-1}-D_{2}^{-1}\Vert \right)\left\Vert A\right\Vert .
$$
In particular, if we fix $D_{2}$ with no zero entries on the diagonal and take $D_{1}\rightarrow D_{2}$, then $T_{1}A\rightarrow T_{2}A$.
Since you care about the spectral norm, I should also point out that for any diagonal $D=(d_{ij})$,
$$
\left\Vert D\right\Vert _{2}=\max_{i}\left|d_{ii}\right|.
$$
You can apply this fact to the previous inequality.
