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@Dave’s example works, for the reason he gave, if you replace $\mathbb{Z}_p$ by $\mathbb{Q}_p$. $\text{Hom}_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Q}_p)\cong\mathbb{Q}_p$, whose cardinality is the continuum $\mathfrak{c}$. But since $\mathbb{Q}_p$ is a vector space over $\mathbb{Q}$ of dimension $\mathfrak{c}$, $\text{Hom}_\mathbb{Z}(\mathbb{Q}_p,\mathbb{Q}_p)$ has cardinality $2^\mathfrak{c}$.
@DavidWhite To be fair, the OP posted a different question (without the “finite”) which got several answers. They then edited the question, adding “finite” and got a couple more answers to the new question. I rolled back to the original question, as I didn’t think it was right to change the question after getting good answers. The OP was then advised to post the revised question as a new question, which they have now done.
The original question seems to have had four good answers. It was then edited to specify finite (co)products. I don’t think changing the question like that, after people have put in effort to answer the original question, is right. So I have rolled back to the original question.
