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Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?

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For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{\text{simp}}(EG, \mathbb{Z})$ to the chain complex with $\mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{\text{simp}}_1(EG, \mathbb{Z})=\mathbb{Z}[G]$.

For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes $\require{AMScd} \begin{CD} \ldots @>>> 0 @>>> \mathbb{Z} @>\times 2>> \mathbb{Z} @>>> 0 @>>> \ldots\\ @. @| @VV0V @VV\text{mod }2V @| @.\\ \ldots @>>> 0 @>>> 0 @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \ldots \end{CD}$ are homologous, but after tensoring with $\mathbb{Z}/2\mathbb{Z}$ they are not. Hence they cannot be homotopic.

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