As in the title, for a category with only one object $\mathcal C$, $\mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $\mathcal C$ and $\mathcal C^{op}$ play different roles?
Thank you in advance.
On
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BM\to Set$ can be identified with a left $M$-set, while a functor $BM^{\mathrm{op}}\to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is $$M=\langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=b\rangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{\mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
$\mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $\hom(\cdot,\cdot)$. Let us write the multiplication by $\circ$. Then $\mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $x\circ' y=y\circ x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.