Question: Let $n$ be a positive integer and considere the ring $\mathbb Z\_n$ of the residual classes modulo $n$. My intuition tells me that there is no two distinct ideals of $\mathbb Z\_n$ with the same number of elements. Is this true?

Answer: 
If $I$ is an ideal of $\mathbb{Z}\_n$ whit $d$ elements, then it is generated by $n/d$ i.e $I=(n/d)$, hence the uniqueness.
An ideal of $Z\_n$ is the same thing as a subgroup of $Z\_n$. The group $Z\_n$ is the set of all $\bar{x}$, where $x \in \mathbb{Z}$. We have that $\bar{x} = \bar{x'}$ if and only if $x \equiv x' \pmod n$.

Let $H$ be any nontrivial subgroup of $Z\_n$. Let $y \in \mathbb{N}$ be the least positive integer such that $\bar{y} \in H$. 

**Lemma: $\bar{y}$ generates $H$.**

The subgroup $S$ generated by $\bar{y}$ is the set $k \bar{y}$, where $k$ runs through all the integers. Already we have $S \subseteq H$, so let $\bar{a} \in H$. Then, we can divide $a$ by $y$, i.e. we can find integers $q, r$ such that $a = qy + r$ and $0 \leq r \leq y - 1$. If $r = 0$, then $$a = qy \Rightarrow \bar{a} = q \bar{y}$$ which means $\bar{a} \in S$. Suppose, however, $1 \leq r \leq n - 1$. Now $\bar{a} = \overline{qy + r} = q \bar{y} + \bar{r}$, so $\bar{r} = \bar{a} - q \bar{y} \in H$ (because $\bar{a}$ and $q \bar{y}$ are $\in H$). But this is a contradiction, since we had supposed that $y$ was the smallest positive integer such that $\bar{y} \in H$. Hence $H \subseteq S$. $\blacksquare$

Using this lemma you should be able to prove your claim.