Can we find $\gamma>0$ such that $ \|x-A^{-1} y\|_2 \leq \gamma \|x-y\|_2$?

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Let $x, y \in \mathbb{R}^d$, and $A \in \mathbb{R}^{d \times d}$ a positive definite matrix such that $\alpha I \preceq A \preceq \beta I$ where $ 0 < \alpha< \beta$ and $I$ is the identity matrix of order $d$. Can we find $\gamma > 0$ such that we have the following inequality:

$$ \|x-A^{-1} y\|_2 \leq \gamma \|x-y\|_2$$

where $\|\cdot\|_2$ is the Euclidean norm.

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It is not possible to have the existence of $\gamma$ such that the inequality holds for all $x,y$: Take $y$ such that $y\ne A^{-1}y$ and set $x:=y$. Such $y$ is guaranteed to exist if $A\ne I$.

If you want this just for a fixed pair $x\ne y$, then $\gamma := \|x-A^{-1}y\| / \|x-y\|$ works.