= Solution
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 actions>
$$
g\cdot k=q(g)k,\qquad h\cdot k=kr(h)^{-1}.
$$
The inverse is essential for the right multiplication formula to define a left <group action>: $h_1\cdot(h_2\cdot k)=k r(h_2)^{-1}r(h_1)^{-1}=(h_1h_2)\cdot k$. 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 is
$$
\boxed{A^G/H=\varnothing\longrightarrow(A/H)^G=\{*\},}
$$
which 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>.
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