I am struggling with exercise 6.12 of Chapter III from Barnsley's Fractals Everywhere, 2nd edition. The exercise is as follows:
Let $(X, d)$ be a compact metric space. Let $f : X \to X$ have the property $\lim\limits_{n \to \infty} f^{\circ n }(X) = x_f$ . Find a metric $\widetilde{d}$ on X such that $f$ is a contraction mapping, and the identity is homeomorphism from $(\overline{X}, d) \to (\overline{X}, \widetilde{d}).$
I copied it verbatim but I assume that $\{x_f\}$ should be used instead of $x_f$ and that $\overline{X} = X$
Solution to more general problem can be found in paper by Mayers (Corollary 1.1 applies to any compact space). It is a bit tricky though.
Perhaps in this situation there is a simpler solution, but it doesn't seem to be proposed
$$\tilde d(x, y) = \sum_{n = 0}^\infty \frac {d(f^{\circ n}(x),f^{\circ n}(y)) } {2^n}. $$
(while this is an equivalent metric, we have $$\tilde d(f(x), f(y)) = \sum_{n = 0}^\infty\frac {2 \cdot d(f^{\circ n+1}(x),f^{\circ n+1}(y)) } {2 \cdot 2^n} \leq 2 \tilde d(x, y), $$ and we can't really make it work with any constant less than $1$ this way)
I'll put my failed approach here as well. If we knew, that for some $0 < \lambda < 1$ sequence $\lambda^{-n} d(f^{\circ n}(x), x_f)$ was bounded for all $x$, we could use $$ \hat d(x, y) = \sup_{n \geq 0} \lambda^{-n}d(f^{\circ n}(x), f^{\circ n}(y)), $$ (but that is a much stronger condition).