Since the unit group of $(\mathbb{Z}/n\mathbb{Z})$ is not cyclic, then $n$ is not prime, an odd prime power, twice an odd prime power, or $4$.
If $n$ is a power of $2$, $n=2^{m}$, with $m\gt 2$, then $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is isomorphic to $C_2\times C_{2^{m-2}}$, which has exponent $2^{m-2}$. Since $\phi(n) = 2^{m-1}$ is a multiple, the conclusion follows.
Otherwise, there are primes $p\lt q$ that divide $n$. Write $n$ as $p^a q^b r$, where $\gcd(p,r)=\gcd(q,r)=1$. Then $\phi(n) = (p-1)p^{a-1}(q-1)q^{b-1}\phi(r)$. Thus, $(\mathbb{Z}/n\mathbb{Z})^{\times} \cong (\mathbb{Z}/p^a\mathbb{Z})^{\times}\times (\mathbb{Z}/q^b\mathbb{Z})^{\times}\times (\mathbb{Z}/r\mathbb{Z})^{\times}.$ Now, if both $p$ and $q$ are odd, then $\phi(p^a)$ and $\phi(p^b)$ both divide $\phi(n)/2$; so does $\phi(r)$, so the group is of exponent $\phi(n)/2$.
If $p=2$ and $a=1$, then $r$ is odd and greater than $1$, so both $(q-1)q^{b-1}$ and $\phi(r)$ are divisors of $\phi(n)/2$. If $p=2$ and $a\gt 1$, then $2^{a-1}$ divides $\phi(n)$, and $\phi(q)$ and $\phi(r)$ are both divisors of $\phi(n)/2$, and the exponent of $(\mathbb{Z}/2^{a}\mathbb{Z})^{\times}$ is $2^{a-2}$, which is also a divisor of $\phi(n)/2$. So $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is of exponent $\phi(n)/2$.