Let $A,B$ be two $n\times n$ invertible matrices with complex entries.
Assume that $B$ is very close to the identity matrix, i.e., $\|B-I\|\ll1$.
Can there be a bound on $|\|ABA^{-1}\|-1|$ uniform in $\|A\|$?
Of course, we have $\|ABA^{-1}\|\leq1+\|A(B-I)A^{-1}\|\leq1+\|A\|\|A^{-1}\|\|B-I\|$.
But if $\|A\|$ is very large, then this becomes problematic. Since $B$ is almost the identity matrix, it feels like there should be a way to get a bound that exploits the factors of $A$ and $A^{-1}$.
Ideally provide a proof or a counter example.
No, there can't be a bound that only depends on $\|A\|$. The problem isn't when $\|A\|$ is large, it's when $\|A^{-1}\|$ is large (which can happen with $\|A\|$ staying bounded).
Try $$B = \pmatrix{1 & \epsilon\cr 0 & 1\cr}, \ A = \pmatrix{\epsilon^2 & 0\cr 1 & 1\cr} $$ Then $$ A B A^{-1} = \pmatrix{1-\epsilon & \epsilon^3\cr -1/\epsilon & 1 + \epsilon}$$ so $\|A B A^{-1}\| \to \infty$ as $\epsilon \to 0$.