Suppose $\mathsf C$ is an algebraic category/variety, i.e a category of some set-based models of an algebraic theory and their homomorphisms. Suppose $\mathsf C$ is pointed and has kernels. Suppose moreover the theorem $A/\operatorname{Ker}f\cong \operatorname{Im}f$ holds for each $f:A\to B$, where the image is the set theoretic image with the induced structure.
What's an example of such a $\mathsf C$ for which the fibers of a homomorphism are not in bijection with each other? I think Heyting algebras might be an example but I keep getting confused in trying to work out an example.
Using Kevin's suggestion, consider Heyting semilattices (they have the signature $\top, \wedge, \Rightarrow$), let $A = \{\bot, \top\}$ be the three-element and $B = \{\bot,1/2, \top\}$ with the structure defined as in this article.
Then define $f : A\to B$ by $f(\bot) = \bot, f(1/2) = f(\top) = \top$. You can verify that this is a morphism of Heyting semilattices, although: $f^{-1}(\bot) = \{\bot\}$ and $f^{-1}(\top) = \{1/2, \top\}$.
An algebraic theory, such that the equivalence classes of congruences are all in bijection is called "congruence uniform". In a theory with a unique constant $1$ (i.e. pointed) containing two binary operations $\cdot$ and $/$ such that: $$1\cdot x = x\cdot 1 = x, (x\cdot y)/y = x, (x/y)\cdot y = x$$ this property always holds (see "Mal'cev, protomodular, homological and semi-abelian categories" 5.32, , 5.3.4, 5.3.7 and 5.3.10). In particular this is the case, if the theory contains a group operation (so for groups, rings, modules,...)