$\lambda(n^2)$ versus $n\lambda(n)$ Let $\lambda$ be the Carmichael function.
What is the relationship between $\lambda(n^2)$ and $n\lambda(n)$ ?
It is easy to prove that $\lambda(n^2)\le n\lambda(n)$ $\ (\star)$.
Actually, $\lambda(n^2)$ divides $n\lambda(n)$.
Is there anything else to be said?
In particular, for what $n$ do we have $\lambda(n^2)=n\lambda(n)$ ?
(For Euler $\phi$, we have $\phi(n^2)=n\phi(n)$, but $\phi$ is multiplicative while $\lambda$ isn't.)

$(\star)$
  Take $x$ with $(x,n^2)=1$. Then $(x,n)=1$ and $x^{\lambda(n)}\equiv 1 \bmod n$. Write $x^{\lambda(n)}=1+an$. Then $x^{n\lambda(n)}=(1+an)^n=1+\binom{n}1an+ \binom{n}2 (an)^2+\cdots+(an)^n \equiv1 \bmod n^2$. Thus, $n\lambda(n) \ge \lambda(n^2)$, because $\lambda(n^2)$ is the least exponent for $\mathbb{Z}_{n^2}^*$.

 A: I doubt there is a simple relationship. But based on your proof for $\lambda(n^2)\mid n\lambda(n)$ I found a more understandable description of the quotient $\frac{n\lambda(n)}{\lambda(n^2)}$. Let $g_n=\gcd\{x^{\lambda(n)}-1:\gcd(x,n)=1\}$ so $n\mid g_n$. Certainly $\lambda(n^2)\mid\frac{n^2\lambda(n)}{\gcd(n^2,g_n)}$ because for all $x$ coprime to $n$, $x^{\frac{n^2\lambda(n)}{\gcd(n^2,g_n)}}=(1+a_xg_n)^{\frac{n^2}{\gcd(n^2,g_n)}}=1+a_xg_n\frac{n^2}{\gcd(n^2,g_n)}+\ldots\cdot g_n^2\equiv1\pmod{n^2}$. On the other hand, $1\equiv x^{\lambda(n^2)}=(x^{\lambda(n)})^{\frac{\lambda(n^2)}{\lambda(n)}}=(1+a_xg_n)^{\frac{\lambda(n^2)}{\lambda(n)}}=1+a_xg_n\frac{\lambda(n^2)}{\lambda(n)}+\ldots\cdot g_n^2\pmod{n^2}$ where for some $x$, $\gcd(a_x,n)=1$. For such $x$ we find $n^2\mid\frac{g_n\lambda(n^2)}{\lambda(n)}$, hence $n^2\mid\frac{\gcd(n^2,g_n)\lambda(n^2)}{\lambda(n)}$. Combining both divisibility relations gives $\frac{n\lambda(n)}{\lambda(n^2)}=\frac{\gcd(n^2,g_n)}{n}$. That is, the non-trivial part of $g_n$ which divides $n^2$.
