Let $\mathcal{C}$ be the full subcategory of $\textbf{Ab}$ whose objects are finitely generated abelian groups. Let $F:\mathcal{C}\rightarrow\mathcal{C}$ be the functor sending $A\in\mathcal{C}$ to $A_{\text{tor}}\oplus A/A_{\text{tor}}$. I want to show there is not natural isomorphism $F\rightarrow\text{id}_{\mathcal{C}}$.
For the sake of contradiction, suppose there exists a natural isomorphism $u:F\rightarrow\text{id}_\mathcal{C}$. For each $A\in\textbf{Ab}$, let $v(A):=\iota_{A,2}\circ\pi_A$ where $\pi_A:A\rightarrow A/A_{\text{tor}}$ is the canonical surjection and $\iota_{A,2}:A/A_{\text{tor}}\rightarrow A_{\text{tor}}\oplus A/A_{\text{tor}}$ is the canonical injection. Clearly $v:\text{id}_{\mathcal{C}}\rightarrow F$ is a natural transformation. Thus we have $u\circ v:\text{id}_{\mathcal{C}}\rightarrow\text{id}_{\mathcal{C}}$ and so there exists $n\in\mathbb{Z}$ such that
$$u(A)\circ v(A)=n\cdot\text{id}_A$$
for all $A\in\mathcal{C}$. In fact $n=u(\mathbb{Z})(v(\mathbb{Z})(1))=u(\mathbb{Z})(0,\pi_{\mathbb{Z}}(1))$.
To get a contradiction, it's sufficient to show that necessarily $n=\pm1$. How to prove this?
Edit: Note that $A_{\text{tor}}$ is the torsion subgroup of the abelian group $A$.
Best Answer
There is a detailed solution of this in E. Riehl's Category Theory in Context (Proposition $1.4.4$).
You are almost there. To finish the proof we consider the components of the constructed natural transformation at both $A=\mathbb Z$ and $A=\mathbb Z/2n\mathbb Z$. From the first case we get that $n\ne0$ while the second case implies that $n=0$ as the given component factors through the trivial group ($\mathbb Z/2n\mathbb Z$ is torsion). This completes the argument.