For what integers $N$ does $\phi(N^2)=\phi(N)^2$?

Let $N$ be a positive integer and suppose that \begin{equation*} \phi(N^2)=\phi(N)^2 \end{equation*} where $\phi$ is the Euler function. What can we say about $N$? Here is my attempt:

Of course, this equality holds for the trivial case $N=1$ because $\phi(1)=1$. If $N>1$ is a prime integer, the equality cannot hold because in that case \begin{equation*} \phi(N^2)=N(N-1)>(N-1)^2=\phi(N)^2 \end{equation*} Thus, if $N$ is a positive integer such that $\phi(N^2)=\phi(N)^2$, then $N$ is not prime. So, consider its factorization \begin{equation*} N=\prod_{i=1}^n p_i^{k_i} \end{equation*} where the $p_i$s are different prime integers. Then, \begin{equation*} N^2=\prod_{i=1}^n p_i^{2k_i} \end{equation*} and \begin{equation*} \phi(N^2)=\prod_{i=1}^n \phi(p_i^{2k_i})=\prod_{i=1}^n p_i^{2k_i-1}(p_i-1) \end{equation*} while \begin{equation*} \phi(N)=\prod_{i=1}^n p_i^{k_i-1}(p_i-1) \end{equation*} and consequently, \begin{equation*} \phi(N)^2=\prod_{i=1}^n p_i^{2k_i-2}(p_i-1)^2 \end{equation*} Thus, we have the following equality \begin{equation*} \prod_{i=1}^n p_i^{2k_i-1}(p_i-1)=\prod_{i=1}^n p_i^{2k_i-2}(p_i-1)^2 \end{equation*} By simplifying a bit, we obtain \begin{equation*} \prod_{i=1}^n p_i^{-1}(p_i-1)=1 \end{equation*} that is, \begin{equation*} \prod_{i=1}^n \frac{p_i-1}{p_i}=1 \end{equation*} However, I cannot imagine any finite sequence of primes $p_1,\dots,p_n$ having this property...I think I have made something wrong. Can you give me any hint?

Thanks for your help (as always)!