Is every infinity-category equivalent to a category? To a weak 2-category?

As far as I understand it, the composition of morphisms in an $\infty$-category $\cal C$ does not have to be unique, so $\cal C$ does not have to be a category. Is $\cal C$ equivalent to a category (where $\cal C$ is trivially an $\infty$-category)?

Can the $\infty$-structure be straightened to a weak $2$-category (with the trivial higher $\infty$-category structure)?

EDIT: I am confused about this, because $\infty$-categories seem often considered as categories. For example, I saw a statement saying that every $(\infty,0)$-category is an $\infty$-groupoid (and hence a category). Does invertibility of all morphisms imply that $\cal C$ is a category?

Certainly not. The homs $\mathrm{Hom}(a,b)$ between two objects in an $\infty$-category can be modeled by topological spaces or by Kan complexes, and an equivalence of $\infty$-categories induces, in particular, homotopy equivalences between each corresponding pair of homs. In a category, these homs are, of course, mere sets, so any $\infty$-category containing any higher homotopy groups in any of its hom-spaces is not equivalent to a category (and conversely.) If this were possible, then these things wouldn't get anywhere close to as much attention as they do.

An $\infty$-category in which all morphisms are invertible is, again, equivalent to a groupoid if and only if the morphism spaces have no higher homotopy groups.

Instead, an $\infty$-category can be most strictly presented as a simplicially enriched category: the composition is associative on the nose, but you still have to carry around higher simplices in the hom-simplicial sets and think about how to compose them as well. And even then, only very special and quite complicated simplicial categories will have the property that every map between the $\infty$-categories they model arises from a functor of simplicial categories.