For a metric space $(X,d)$, let $\DeclareMathOperator{\Iso}{\operatorname{Iso}} \Iso(X, d)$ be a set of all isometries on $X$. (Function $f:X \rightarrow X$ is isometry if $d(f(x), f(y)) = d(x, y)$, for all $x,y \in X$).
Is there an example of compact metric space $(X,d)$ such that $X$ is uncountable set and $\operatorname{card}(\Iso(X,d)) = \aleph_0$?
No, there is no such example. In fact we do not even need to assume $X$ is uncountable.
$\newcommand{\isom}{\operatorname{Isom}}$ A quicker way to prove this than the original proof below is to observe that by Arzelà–Ascoli $\isom(X)$ is compact. If it is infinite, then it is not discrete, and then since the $\sup$ metric is invariant under postcomposition by isometries (i.e, $d(f\circ g_1,f\circ g_2)=d(g_1,g_2)$), $\isom(X)$ has no isolated points* (see clarification below).
Since it is also compact, hence complete, $\isom(X)$ is perfect, and therefore uncountable (and in fact has cardinality at least $2^{\aleph_0}$).
To see this, first note that if $X$ is compact, the isometry group $\isom(X)$ is a compact subset of the space $\mathcal C(X,X)$ under the uniform metric $d(f,g)=\sup_{x\in X} d(f(x),g(x))$, by Arzelà–Ascoli.
It follows that if $\isom(X)$ is infinite, it cannot be discrete, and thus has members arbitrarily close to the identity $I$.
We may therefore pick a sequence $f_n\in \isom(X)$ with $f_n\neq I$ and $d(f_{n+1},I)\leq \frac{1}{4}d(f_n,I)$ for each $n$.
Now for a finite subset $S=\{n_1,\dots, n_k\}\subseteq \mathbb N$, with $n_1< \dots <n_k$, define $f_S=f_{n_k}\circ\dots\circ f_{n_1}$, and then for arbitrary subsets $S\subseteq \mathbb N$, let $f_S= \lim_{n\to\infty} f_{S\cap\{1,\dots,n\}}$.
It is easy to see that each $f_S$ is an isometry. Moreover, for each $k$ we have $$d(f_{S\cap\{1,\dots,k\}},f_S)\leq \sum_{n=k+1}^{\infty} d(f_n,I)\leq \sum_{n=1}^\infty \frac{1}{4^n}d(f_k,I)=\frac{1}{3} d(f_k,I).$$
Then if $S\neq T\subseteq\mathbb N$, and $k$ is the first integer for which $S$ and $T$ differ, then
\begin{align*} d(f_S,f_T) &\geq d(f_{S\cap\{1,\dots,k\}},f_{T\cap\{1,\dots,k\}})- d(f_{S\cap\{1,\dots,k\}},f_S) - d(f_{T\cap\{1,\dots,k\}},f_T)\\ &\geq d(f_{S\cap\{1,\dots,k\}},f_{T\cap\{1,\dots,k\}})- \frac{2}{3} d(f_k,I)\\ &=d(f_k,I)-\frac{2}{3}d(f_k,I)\\ &=\frac{1}{3}d(f_k,I)>0. \end{align*}
It follows that every distinct subset $S\subseteq \mathbb N$ gives rise to a distinct isometry $f_S$, so the isometry group is uncountable.
Throughout both answers above, I tacitly assumed we were talking about surjective isometries, whereas in the original question isometries were not defined to be a priori surjective. However, this is not a problem, since when $X$ is compact isometries are always surjective.
Since the uniform metric is preserved by post-composition of isometries, for each $f\in \isom(X)$ the map $g\mapsto f\circ g$ is an isometry of $\isom(X)$, and a fortiori is a homeomorphism, which therefore takes isolated points to isolated points. Thus if there is a single isolated isometry $h\in \isom(X)$, then for each $f\in\isom(X)$ the map $g\mapsto f\circ h^{-1}\circ g$ is a homeomorphism taking $h$ to $f$, so $f$ is isolated.
We conclude that $\isom(X)$ either has no isolated points, or is discrete. Since the latter is impossible for an infinite compact set, we have no isolated points.
Note also that this isn’t really about isometries in particular, and is true for any topological group, but I have avoided going into that discussion in case you have not yet been exposed topological groups.
Here is one example. Let $X \subset \mathbb{R}^2, X = \{(x,y): (x,y)\in \mathbb{R}^2, xy = 0, |x| \leq 1, |y| \leq 1\}$. This set is also compact. Let $d$ be the Euclidean metric. Obviously, $X$ is uncountable, but there are only eight isometries, corresponding to the rotations by multiples of $\pi/2$ angles around the origin, perhaps combined with a reflection around the origin. Sets like $Y = \{(x,y) \in \mathbb{R}^2: |x| \leq 1, |y| \leq 1\}$ also work.
