How do you describe congruence classes in general?

If I understand correctly, $C_n$ here is a unary algebra, having a single (cyclic) operation $s=x\mapsto x+1$ mod $n$.

If we have an algebraic structure $A$, then a congruence relation $\theta$ on $A$ is defined as an equivalence relation (i.e. it's a subset of $A\times A$ satisfying reflexivity, symmetry and transitivity) which 'behaves like equality', i.e. also respects all given operations of $A$ in the sense that whenever $\mu:A^n\to A$ is an $n$-ary operation, and $a_1,\dots,a_n,\ b_1,\dots,b_n\ \in A$ such that $a_i\,\theta\, b_i$ (meaning $(a_i,b_i)\in\theta$) for each $i=1,\dots,n$, then also $$\mu(a_1,\dots,a_n)\ \theta\ \mu(b_1,\dots,b_n)\,.$$ (In other words, $\theta$ is also a subalgebra of $A\times A$.)

Congruence relations can be generated by a given pair $(a,b)$ of elements or by a given set $S$ of pairs, i.e. any relation $S\subseteq A\times A$ determines a congruence relation, namely the intersection of all congruence relations (as subsets of $A\times A$) which contain $S$.

Back to $C_n$, take for example $n=6$ and let's figure out the congruence relation $\alpha$ generated by the pair $(1,3)$.

By symmetry, we can deduce $3\,\alpha \,1$.
Applying the condition on the operation $s$ of $C_6$, we get that $1\,\alpha\, 3$ implies $$2=s(1)\ \alpha\ s(3)=4$$ that is $(2,4)\in\alpha$. Which further implies $3\,\alpha\, 5$, then $4\,\alpha\,0,\ 5\,\alpha\,1$ and $0\,\alpha\,2$.

Using symmetry and transitivity, we see that $0,2,4$ are all in $\alpha$ relation with each other and similarly are $1,3,5$. These are the two equivalence classes now. You can verify that this partition indeed defines a congruence relation, and as each pair of it was forced, it must be the smallest congruence relation containing $(1,3)$.

Try to find the congruence generated by $(2,5)$. How many equivalence classes will that have?

Now if a congruence relation $\beta$ is given on $C_n$, let $b$ the smallest among $1,2,\dots,n-1$ such that $0\,\beta\,b$.
Then try to show that $b$ divides $n$ and $\beta$ is generated by the pair $(0,b)$ and specifically, $$x\,\beta\,y\ \iff\ b\,|\,(y-x)\,.$$ Finally, it's possible that such a $b$ doesn't exist (when $0$ is only $\beta$-related to itself). Prove in that case $\beta$ is the equality relation (aka 'diagonal') $\{(a,a):a\in A\}$.