In my lecture notes on differential geometry I stumbled upon the following statement regarding the euclidean metric $d$ restricted to the unit sphere $\mathbb S^{n-1}$ and the angle-distance $d_S$ on $\mathbb S^{n-1}$.
The metrics $d$ and $d_S$ are equivalent, not just in the weak sense of inducing the same topology, but in the strong sense that the isometries of $\mathbb S^{n-1}$ are the same with respect to either metric, since the two metrics are simply monotonic functions of one another.
My question is whether it is true in general that if two metrics $d,d'$ on a set $X$ satisfy $cd\leq d'\leq Cd$ for some $C\geq c>0$, then the isometry groups $\operatorname{Isom}_d(X)$ and $\operatorname{Isom}_{d'}(X)$ coincide?
Here is my incomplete attempt at proving this. For sake of reducing unneeded parentheses, I'm gonna write $fx:=f(x)$ for maps $f\colon X\to X$ and $x\in X$.
Let $f\in\operatorname{Isom}_{d'}(X)$, i.e. $d'(fx,fy)=d'(x,y)$ for all $x,y\in X$. It immediately follows that $$c\max\{d(x,y),d(fx,fy)\}\leq\underbrace{d'(fx,fy)}_{=d'(x,y)}\leq C\min\{d(x,y),d(fx,fy)\}.$$ We can distinguish two cases: $d(x,y)\leq d(fx,fy)$ or $d(fx,fy)\leq d(x,y)$. In the former case the previous inequality reduces to $cd(fx,fy)\leq Cd(x,y)$, which does not help all that much, because $C/c\geq1$. The latter case is no help either, and it does not look like I can get more out of those inequalities.
Do you have any idea how to prove this claim (if it is true in the first place)?