Okay so I was kind of winging it, especially in the second direction. I wanted to make sure what I wrote is not complete nonsense..
Assume $f$ is injective.
Since $R$ is an equivalent relation, it must be reflexive, symmetric and transitive.
Therefore in order for f to be injective, each element in A must only relate to itself, since if it relates to a different element in A, it must also relate to itself and the function will no longer be injective. Since every element must only relate to itself, there are exactly n equivalence classes, since there are n elements in $A$ and therefore the size of the quotient set is $n$.
Assume there are n quotient sets.
Since $R$ is an equivalent relation, the maximum size of the quotient set of $A$ by $R$ is $|A|$, and in this case every element relates only to itself.
Therefore since $|A| = n$, and the the size of the quotient set is n, it follows for every $a,b ∈ A \ \ aRb$ iff $a = b$. Since every element is only related to itself, then $f$ is injective.