I originally posted this question to MSE but there were no answers except for a partial one from me, so I'm trying again here. I'm an undergrad, not a researcher, so forgive me (and correct me!) if my terminology is nonstandard, or my ideas confused.

By "induction" I mean "no proper subalgebras". My thinking goes like this:

- For natural numbers, recursion and induction are in some sense the same thing. In particular, given a recursive definition of $f$ you would prove its totality roughly by saying "if I can define $f$ on $1\dots n$, then I can define it on $n+1$", i.e. by induction.
- The proper categorical notion of recursion is initial algebras – in particular, for $F(X) = 1\sqcup X$ an initial algebra is a natural number object, the property of being initial being precisely what you need to define functions by recursion.
- An initial algebra automatically has a notion of induction: initial objects have no proper subobjects, so if you have some subobject that is closed under the algebra operations, this means precisely that the inclusion is an algebra homomorphism, and therefore an isomorphism.
- I'd really like to go the other way, but in general the implication "$I$ is initial $\implies$ every mono into $I$ is an iso" cannot be reversed (its dual has a counterexample in $\mathbf{Set}$, in that every epi into $0$ is an iso but $0$ is not terminal).
Are there circumstances where an algebra having no proper subalgebra means it is initial?

Observations that have been made since posting the question:

- There is an algebra $1 \sqcup \mathbb Z/n\mathbb Z \to \mathbb Z/n\mathbb Z$ that has no proper subalgebras, so induction works for it. Nevertheless it is not initial. The problem is that the algebra satisfies equations, since $n$ successors give the identity, so there aren't morphisms to algebras that don't satisfy those equations. This suggests that the algebra needs to be "free" in some sense.
- In light of the above, I might decide I'm interested in
*quasi*-initial objects, since it is at least true that any morphism from $\mathbb Z/n\mathbb Z$ that is an algebra homomorphism is unique. It turns out that in any category with equalisers, if an object has no proper subobjects then it is quasi-initial, and of course a category of algebras has equalisers if its underlying category does.

What's the "freeness" condition that I want? What can I do to show that *any* morphisms exist from an object with no proper subobjects?