Let $(X, d)$ be a compact metric space.

We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-

**covering**of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, x_i) \leq \varepsilon$. Let $$ \operatorname{Cov} (X, \varepsilon) := \min \{n: \exists \varepsilon \text {-covering of } X \text { with size } n\} $$ be the $\varepsilon$-covering number of $X$.We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-

**packing**of $X$ if $d(x_i, x_j)>\varepsilon$ for all distinct $i, j$. Let $$ \operatorname{Pack} (X, \varepsilon) := \max \{n: \exists \varepsilon \text {-packing of } A \text { with size } n\} $$ be the packing number of $A$.

Let $(X, d)$ and $(X', d')$ be metric spaces. The spaces $X$ and $X'$ are said to be isometric (denoted by $X \cong X'$) if there is a bijective isometry between them.

@Noam gave below example in his answer:

For $\delta \in (0, 2]$, let $E_\delta$ be the metric space consisting of three points $A,B,C$ with $d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$. Then for all $\delta, \delta' \in [1, 2]$, $$ \operatorname{Cov} (E_\delta, \varepsilon) = \operatorname{Cov} (E_{\delta'}, \varepsilon) = \begin{cases} 3 & \text{if} \quad \varepsilon < 1, \\ 1 & \text{if} \quad \varepsilon \ge 1. \end{cases} $$

This example shows that $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ for all $\varepsilon>0$ does not necessarily imply $X \cong X'$. Back to @Noam example, it's clear that $$ \operatorname{Pack} (E_1, 1) = 0 \neq 2= \operatorname{Pack} (E_{2}, 1). $$

I would like to ask if below statement is true, i.e.,

If $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ and $\operatorname{Pack} (X, \varepsilon) = \operatorname{Pack} (X', \varepsilon)$ for all $\varepsilon>0$, then $X \cong X'$.

Thank you so much for your elaboration!