Assume $R$ is an equivalence relation over $A$ and there is a bijection between $A$ and $A / R$. Does this entail $R = \left\{ (x, y ) \in A^2 : x = y \right\} $? What I thought is the following.
Assume $A$ is an infinite set. Then we can find a counter-example. For example, if $A = \mathbb{N}$ and $R$ the equivalence relation s.t. $A / R$ is the partition
$$\left\{ \left\{ 1 \right\}, \left\{ 2, 3\right\}, \left\{ 4, 5, 6 \right\}, \left\{ 7, 8, 9, 10 \right\}, \ldots \right\} $$
Then $F(1) = \left\{ 1 \right\} , F(2) = \left\{ 2, 3 \right\}, F(3) = \left\{ 4, 5, 6 \right\}, \ldots $ is a bijection.
However, if $A = \left\{ a_1, \ldots, a_n \right\} $ a finite set, and $F$ is bijective, we must have
\begin{align*} F(a_1) = X_1, \ldots, F(a_{n}) = X_{n} \end{align*}
with $X_i \neq X_j$ for $i, j \in [1, n]$. This entails $|A / R| = |A|$, which implies $A / R$ is a partition of $A$ into singleton sets. In other words,
$$A / R = \left\{ \left\{ a_1 \right\}, \ldots, \left\{ a_n\right\} \right\} \Rightarrow R = \left\{ (x, y) \in A^2 : x = y \right\} $$
So, the answer to the question seems to differ with regards to whether $A$ is finite or infinite. Is this correct?
You've got it right. We have $|A| = |A/R|$ (two sets have the same cardinality iff there's a bijection between them), and any surjective function between two finite sets of the same cardinality is a bijection. Thus the surjective function $v : A \to A/R$ that maps each element of $A$ to its partition is also injective, and hence no two elements are in the same partition.