$n\in\mathbb{N}$ such that for every divisor $d$ of $\phi(n)$ there exist a element with order $d$ modulo $n$

Question

I am looking to describe the set of $n\in\mathbb{N}$ such that for every divisor $d$ of $\phi(n)$ there exist a element with order $d$ modulo $n$.

I noted already that the $n$ needs to admit a primitive root modulo $n$, so by that I have already that $n\in\{2,4,p^k,2p^k | p$ is an odd prime, $k\geq 1\}$.

Finding a counter example in this set is taking a long time so I am looking for hints of where to look or a hint as to what is an element that doesn't work that still has a primitive root.

Answer 1

If $x$ is a primitive root modulo $n$, so that $x^d \not\cong 1 \mod n$ for $0<d<\phi(n)$, then $x^{\phi(n)/d}$ has order $d$ modulo $n$.

Answer 2

Even more - if we denote the Carmichael function $\lambda (n)$ as the least positive integer $t$ that $x^t \equiv 1 \pmod{n}$ for all $(x,n)=1$, we have that there exists an integer $x$ such that $\text{ord}_m(x)=d$ if and only if $d|\lambda (m)$