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Let $q:G\twoheadrightarrow K$ and $r:H\twoheadrightarrow K$ be the two quotient homomorphisms. On the underlying set $A=K$, define the commuting group actionsThe inverse is essential for the right multiplication formula to define a left group action: $h_1\cdot (h_2\cdot k)=kr(h_2{)}^{-1}r(h_1{)}^{-1}=(h_1{h}_2)\cdot k$. Left and right multiplication commute.

Because $q$ is onto and $K$ is nontrivial, there is no fixed point of a group action for $G$: if every $q(g)k=k$, cancellation would make every element of $K$ the identity. Thus $A^G=\varnothing$. Because $r$ is onto, the $H$-action is transitive and $A/H$ is a singleton. The induced $G$-action on that singleton is trivial. Consequently the comparison iswhich is not an isomorphism. This proves the common quotient obstruction to commutation of fixed points and orbits, so limits of shape $G$ do not commute with colimits of shape $H$ in the Category of sets.