In any category, an isomorphism is an arrow $f \colon X \to Y$ together with an inverse arrow $f^{-1} \colon Y \to X$ such that $f \circ f^{-1} = 1_{Y}$ and $f^{-1} \circ f = 1_{X}$.
Now, clearly in some (algebraic) categories like $\mathsf{Group}$, $\mathsf{Set}$, etc. this definition is equivalent to the bijective morphisms. In others like $\mathsf{Pos}$, there are bijective morphisms between non-isomorphic posets.
But do we have equivalence in all small categories? Since any small category is isomorphic to one where the objects are sets and the arrows are functions.
There is no such thing as a "bijective morphism" in a category. The notion of "bijective" only makes sense in a concrete category, which is a category $C$ equipped with the additional data of a faithful functor $U : C \to \text{Set}$, and the meaning of "bijective" depends on the choice of $U$.
The failure of "bijective" morphisms to be isomorphisms has nothing to do with smallness, and any counterexample you've ever seen can be converted into a small counterexample by ignoring all objects other than the two objects involved in the counterexample.
In fact here is a minimal counterexample: take $C$ to be the category $\{ \bullet \to \bullet \}$ consisting of two objects and a morphism $f$ from one to the other, and take $U : C \to \text{Set}$ to be the functor sending both objects to the one-element set and $f$ to the unique function from this set to itself. Then $U(f)$ is a bijection but $f$ is not an isomorphism because there aren't any morphisms in the other direction at all. Here $C$ is not only small but finite.