Given a symmetric matrix $H$ which is already "close to" diagonal. Now a matrix $G$ can be found such that, when multiplied from the left:
$$ G H = I $$
This implies $G=H^{-1}$ and hence multiplying $H$ from the right yields the same result:
$$ H G = H (H^{-1}) = I $$
Now the question is: Suppose $G'$ is found such that it does not perfectly diagonalize, i.e.:
$$ G' H \approx I $$
Can anything be said about $H G'$? In particular, how far off from being diagonal will $G' H$ as compared to $G' H$?
I guess this also requires some sort of "diagonally" measure. For now I would take the L2 norm of the diagonal elements of a matrix M$M divided by the L2 norm of the rest:
$$ \sqrt{\frac{\sum_{i=1}^N |M_{ii}|^2 }{\sum_{i \neq j} |M_{ij}|^2}} $$
but any other meaningful measure would be fine too.
Proposition. Nothing can be said about $HG-I$, except if $H$ is not too badly conditioned.
Proof. Let $G=H^{-1}+U$; then $GH=I+UH$ where $\Delta=UH$ is small. Note that $||U||\leq ||\Delta||||H^{-1}||$ and that, if $H$ has small eigenvalues, then $||H^{-1}||$ may be large and $U$ may be large.
Finally, $HG=I+HU$ and $||HU||\leq ||\Delta||cond(H)$ is not necessarily small (except if $cond(H)$ is "reasonable").
Example: $H=\begin{pmatrix}\epsilon&\epsilon\\\epsilon&1\end{pmatrix},U=\begin{pmatrix}0&0\\1&\epsilon\end{pmatrix}$ where $\epsilon$ is small. $\square$