Suppose $n=pq$ where $p,q$ are prime. A number $z$ is said to be a $n$th-residue modulo $n^2$ if there exists $y\in \mathbb{Z}_{n^2}^*$ such that $z = y^n \bmod n^2$.
It's claimed by this paper that
the set of $n$-th residues is a multiplicative subgroup of $\mathbb{Z}_{n^2}^*$ of order $\phi(n)$.
It's easy to verify that $n$-th residues form a subgroup. But how should one go about calculating its order?
One way to see it is the following (I assume $p \neq q$).
First $\mathbb{Z}_{n^2}^*$ is a group of order $\phi(n^2) = \phi(p^2q^2) = pq(p-1)(q-1) = pq \phi(pq) = n\phi(n)$.
Now consider the group homomorphism $f: \mathbb{Z}_{n^2}^* \to \mathbb{Z}_{n^2}^*$ given by $f(x) = x^n$. We want to compute the size of the image of $f$.
The kernel of $f$ consists of the $x$ so $x^n = 1$ in $\mathbb{Z}_{n^2}^*$. Since $\mathbb{Z}_{n^2}^* \cong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z} \times \mathbb{Z}/(q-1)\mathbb{Z}$ and $n$ is relatively prime to $p-1, q-1$ we see that the kernel has size $n$ (the kernel consists of things of the form $(a, 0, 0)$ for $a \in \mathbb{Z}/n\mathbb{Z}$).
We know that the product of the sizes of the image and the kernel of $f$ is the size of $\mathbb{Z}_{n^2}^* = n\phi(n)$; hence the size of the image of $f$ is $n\phi(n)/n = \phi(n)$.