So my method goes like this: We have 16 ordered pairs. If R is:
Reflexive: It has to include $(1,1), (2,2), (3,3), (4,4)$. So $2$ choices for each of the remaining $12$ pairs.
Symmetric: If it includes $(x,y)$, then it also has to include $(y,x)$, so $2$ choices for each of the $12/2 = 6$ remaining pairs of pairs .
Transitive: This is where it gets murky. I cannot think of a method that counts all possibilities correctly.
We have $2^6 = 64$ reflexive and symmetric relations, and I know that $15$ of them are also transitive (I've seen the Bell number - partitions method), but is there a way to reach the same result with plain old counting?
Since each (directed) graph defines a binary relation: $(x,y) \in R$ iff there is an edge between $x$ and $y$, an alternative is to count graphs with properties of equivalence relation, there are 15 such graphs:
Certain properties are more "natural", for example transitivity means that path on 3 vertices $P_{3}$ implies triangle $K_{3}$. Symmetry means there are loops (not drawn here to reduce clutter) and reflexivity means we can draw undirected graphs since direction in one way implies direction in another.
More alternatives for counting equivalence relations can be found in the answers to similar question here.