Given a non-empty set, A, and an empty relation, R, on that set A, can it be the case that the relation R is an equivalence class?
Transitivity. (a,b) in R, (b,c) in R ===> (c,a) in R. This is trivially true because there is no a, b, c, in R, so the implication is true by way of the antecedent being false.
Symmetry. (a,b) in R ==> (b,a) in R. Again, trivially, the implication is true because the antecedent is false.
But what about reflexivity. If a in R, then a in R. But no a in R -- is this true or false?
An equivalence relation on a non empty set can't be empty, because it's reflexive. So, for any $a\in A$, you have $(a,a)\in R$. Now there is some $a\in A$.
From a slightly different point of view, an equivalence relation on $A$ always contains the identity relation $$ \Delta_A=\{(a,a):a\in A\} $$ which is empty if and only if $A$ is empty.
It's true that the empty relation is transitive and symmetric (also antisymmetric, by the way) on every set.
Note that the main purpose of equivalence relations is to partition the set into pairwise disjoint subsets: $$ [a]_R=\{x\in A: (a,x)\in R\} $$ and so reflexivity (hence non emptyness of $R$) is essential to have $a\in[a]_R$. With the empty relation you can't have a partition defined this way.