Condition on isomorphism to opposite category I've been very interested in when a category $C$ is isomorphic to it's opposite category. Playing around, I conjecture that the following condition is necessary and sufficient for a category $C$ with zero-morphisms:


*

*For all objects $O$ in $C$ and morphisms $f$ where $\operatorname{src}(f)=O$, if both objects exist then $O/\operatorname{ker}(f)$ is isomorphic to $f(O)$.


Where the concepts of images, quotients, kernels are extended via universal properties. 
Is this true? Does every "self-dual" category with zero-morphisms have this condition and is every category with this condition self-dual?
 A: Your condition doesn't characterize self-dual categories, and in fact it is not even self-dual : indeed it holds in the category of groups but not in its dual.
Indeed, the first isomorphism theorem tells you precisely that for any group $G$ and group homomorphism $f:G\to H$, we have an isomorphism $G/\ker(f)\cong f(G)$, so the property holds. Now the quotient $G\to G/\ker(f)$ is the cokernel of the kernel of $f$; thus the dual property would be that the factorisation $G\to \ker(\operatorname{coker}(f))$ of $f$ through the kernel of its cokernel is an epimorphism, and thus surjective. But the cokernel of $f$ is the quotient by the normal closure $\widehat{f(G)}$ of its image; thus if the image of $f$ is not normal in $H$, the factorisation $G\to \ker(\operatorname{coker}(f))$ will be equal to the surjection $G\to f(G)$ followed by a proper injection $f(G)\to \widehat{f(G)}$, and thus it will not be surjective.
In particular, this tells you that even though it satisfies your condition, $\mathbf{Grp}$ is not equivalent to its dual.
