We have:
$$\dfrac{6}{\pi^2}\lt\dfrac{\phi(n)\sigma(n)}{n^2}\le1$$
with equality iff $n=1$.
Are there any known improvements on these bounds?
APPENDUM
For $n$ prime, $\dfrac{\phi\sigma}{n^2}\to1$.
Generally if $n=\prod p_i^{k_i}$, $\dfrac{\phi\sigma}{n^2}=\prod \big(1-\dfrac{1}{p_i^{k_i+1}}\big)$, which means the lower bound is sharp.
Call your function $f$.
Hint 1: If $n$ and $m$ are relatively prime, what is $f(nm)$?
Hint 2: If $n=p_1p_2\ldots p_k$ is a product of primes, what is $f(n)$? Does this expression remind you of anything?