The statement is true; that is:
Let $G$ be a finite group of order $n$. Then either $G$ is cyclic and has exactly $\phi(n)$ elements of order $n$, or else $G$ is not cyclic and has no elements of order $n$.
Proof. $\langle g\rangle$ is a subgroup of $G$ of order $|g|$ for each $g\in G$. By definition, $G$ is cyclic if and only if there exists $g\in G$ such that $\langle g\rangle = G$. So for $G$ of order $n$ we have: $\begin{align*} G\text{ is cyclic} &\iff \text{there exists }g\in G\text{ such that }\langle g\rangle = G\\ &\iff\text{there exists }g\in G\text{ such that }|\langle g\rangle|=|G|\\ &\iff \text{there exists }g\in G\text{ such that }|\langle g\rangle = n\\ &\iff\text{there exists }g\in G\text{ such that }|g|=n. \end{align*}$ Thus, if $G$ is not cyclic, then there are no elments of order $n$. If $G$ is cyclic, then there is at least one element of order $n$, and every element is of the form $g^k$ for some $k\in\mathbb{Z}$, which we may assume satisfies $0\leq k\lt n$ (prove it!). Since $|g^k| = n/\gcd(n,k)$ (prove it!), then the number of elements of order $n$ is the number of integers $k$ such that $0\leq k\lt n$ and $n/\gcd(n,k)=n$, which is the number of integers $k$ such that $0\leq k\lt n$ and $\gcd(n,k) = 1$.