Instead of 1->3, put an arrow from i to j whenever i <= j <= 2i.
I keep forgetting to make a list of what to remember. Peter Johnstone kindly put me out of my misery on this noncategory. Since I don't seem to be having much luck making the example more complicated, maybe making it simpler might work. The ring of integers mod 3 is a one-object monoidal category in the usual way, with multiplication as composition and addition as the monoid. Every arrow is clearly both a square coalgebra and a cubical coalgebra, i.e. we have three of each. Claim: There are no final square coalgebras, but 1 and 2 are final cubical coalgebras. Proof. Square coalgebra homomorphisms f from 2x to x (as square coalgebras) are those that satisfy xf = (2f)(2x). But 2x2 = 1 (mod 3) so every f solves this. Hence there are three square coalgebra homomorphisms from 2x to x, whence no x is a final square coalgebra. Cubical coalgebra homomorphisms f from y to x (as cubical coalgebras) must satisfy xf = (3f)y = 0. But for x other than 0, f = 0 is the only solution. So the two nonzero cubical coalgebras are final. The same example (unless I've forgottten yet another thing) solves the corresponding problem for initial algebras. Vaughan