A natural transformation $`\alpha: F \Rightarrow G`$ must make every square commute: for each arrow $`f: A \to B`$, going down then right ($`G(f) \circ \alpha_A`$) must equal going right then down ($`\alpha_B \circ F(f)`$). Here $`F = G`$ is the "square" functor $`S \mapsto S \times S`$, and $`\alpha`$ is either the flip or the identity. Pick a corner map and an element of $`A \times A`$ to trace both paths.