If $f$ is the exponent of a number $a$ ($\mod p$) and $\phi(f)=2$ then, $a$ and its reciprocal are the only numbers, which belong to this exponent.
$\phi$ is the euler-phi function
Nothing written about $p$ (whether it is prime or not)
$a^f\equiv1\mod p$, such that $f$ is minimal called the exponent of $a$
I mean it is clear that $a$ and $a^{-1}$ have the same exponent, but how to bring $\phi(f)$ into play ?
It is written in the first $6$ lines of the paper below (On the Representation of an Integer as the Sum of Consecutive Integers by Mason, Thomas E.)
Let $f$ be the exponent of $a$. Let $G_f$ denote the set of all classes modulo $p$ with exponent $f$ and let $H_f$ be the classes with exponent a divisor of $f$ that is $b^f \equiv 1 \pmod{p}$.
Then group of invertible classes modulo $p$ is cyclic (when $p$ is prime) of order $p-1$. And $H_f$ is a subgroup with $f$ elements. The number of elements of maximal order is $\phi(f)$. Thus if $\phi(f)=2$, there are exactly two such elements. Since $a$ and $a^{-1}$ are such elements they are those $2$. (Note the two elements are different, as $f$ is not $2$, since $\phi(f)=2$.)
The claim is in general not true for general $p$, though the condition prime could be weakened. But it seems certain it is intended.