How many tensor products of chain complexes are there? Let $Ch$ be the category of nonnegatively-(homologically-)graded chain complexes of abelian groups. Suppose that $(Ch,\boxtimes)$ is a monoidal biclosed structure.

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*Assume that the forgetful functor $(Ch, \boxtimes) \to (Gr,\otimes)$ is strong monoidal, where $Gr$ is the category of nonnegatively-graded abelian groups and $\otimes$ its usual tensor product.

Thus $(A\boxtimes B)_n = \oplus_{i+j = n} A_i \otimes B_j$, equipped with some differential, in an associative and unital way.
Question 1: Can one classify all such monoidal structures $(Ch,\boxtimes)$, up to monoidal equivalence over $(Gr,\otimes)$?
Notes:

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*The usual tensor product $\otimes$ of chain complexes is an example, with $d(x\otimes y) = d x \otimes y + (-1)^{|x|} x \otimes d y$ but not the only one -- e.g. one might define the differential as $d'(x \otimes y) = (-1)^{|y|} d x \otimes y + x \otimes d y$ for a different example.


*If it's easier to work over a field instead of $\mathbb Z$, that would be an interesting start.


*I've singled out nonnegatively-graded chain complexes, but if the situation changes when passing to unbounded chain complexes, that would be interesting.
Question 2: Are all such monoidal structures $\boxtimes$ symmetric monoidal? In how many ways?
 A: Okay, here's a solution when we allow unbounded chain complexes.
First, note that the automorphism group of $Ch$ over $Gr$ is $C_2^\omega$ (or rather $C_2^{\mathbb Z}$ since we've switched to the unbounded case), because swapping the sign of the $n$th differential on all objects of $Ch$ affects neither the underlying graded abelian groups, nor the notion of chain map (and similarly for swapping the sign of $d_n$ for each $n \in S$, where $S \subseteq \mathbb Z$ is an arbitrary subset). And indeed, the two examples of monoidal structures which I gave are related by such an automorphism -- the one where you swap the sign of $d_n : A_n \to A_{n-1}$ whenever $n$ is odd.
Claim: All the monoidal biclosed structures on $Ch$ over $Gr$ are related to the standard one by such an automorphism.
An easy exercise shows that the only nonidentity automorphism of the standard tensor product over $Gr$ is the one which flips the signs of all the differentials, so most of these monoidal structures are different from one another. And apparently they are all symmetric monoidal (in exactly as many ways as the standard monoidal structure -- however many that is).
Proof:
Consider the short exact sequence $S(0) \to D(1) \to S(1)$ (here and in the following, $D(n)$ is the degree $n$ disk and $S(n)$ is the degree $n$ sphere; we take advantage in the following of the fact that these objects have canonical bases which we sometimes silently identify). Tensoring the exact sequence against itself, we get an exact $3\times 3$ grid. Unitality of the monoidal structure tells us that the differentials on $S(0) \boxtimes D(1)$ and $D(1) \boxtimes S(0)$ are canonically isomorphic to the differential on $D(1)$. By looking at the inclusion maps from these objects to $D(1)^{\boxtimes 2}$, we see that the $d_1$ differential on $D(1)^{\boxtimes 2}$ is given by the codiagonal $[1,1]: D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1 \to D(0)_0 \otimes D(0)_0$. This implies that the $d_2$ differential is of the form $[a,-a] : D(1)_1 \otimes D(1)_1 \to D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1$ for some $a \in \mathbb Z$. Conjugating by an automorphism of $Ch$ over $Gr$, we may assume that $a \geq 0$. Quotienting along the maps to $D(1) \boxtimes S(1)$ and $S(1) \boxtimes D(1)$, we see that the differential on the former is given by $a$ and on the latter the differential is $-a$.
I claim that $a = 1$. For if $a = 0$, then $S(1) \boxtimes D(1)$ has a null differential. Tensoring with $S(-1)$, it follows that $D(1)$ has a null differential, a contradiction. On the other hand, if $a > 1$, then there is an exact sequence $0 \to D(2) \to S(1) \boxtimes D(1) \to S(1)/a \to 0$. Again tensoring with $S(-1)$, we obtain an exact sequence $0 \to S(-1) \boxtimes D(2) \to D(1) \to S(0)/a \to 0$. We know all the groups involved here, and because the differential on $D(1)$ is a unit, the differential on $S(-1) \boxtimes D(2)$ would have to be $1/a$, which is of course impossible.
Now just as we used knowledge of the differential on $S(0) \boxtimes D(1)$ to determine the differential on $S(1) \boxtimes D(1)$, we can work our way up to work out that the differential on $S(n) \boxtimes D(1)$ is standard for all $n \geq 0$. A dual argument allows us to work out negative $n$'s as well.
At this point we know the differential on $D(1) \boxtimes D(1)$. It’s the standard one, which splits as a sum $D(2) \oplus D(1)$. We also have canonical isomorphisms $S(n-1) \boxtimes D(1) = D(n)$ and dually $D(1) \boxtimes S(m-1) = D(m)$. Together we can now determine the differential on $D(n) \boxtimes D(m)$. By compatibility with colimits, this determines the entire monoidal structure.

If we stick to just nonnegatively graded complexes, then I have difficulty ruling out the cases $a=0$ or $a > 1$ in the above argument. I would be interested to see the argument pushed through to completion in this case, but I think I have run out of energy to do it myself right now.
A: tldr: (but also, spoilers !) If you impose the further condition that chain complexes be a $Gr$-algebra, then there is a unique monoidal structure - which one of your two options it is will depend on what $Gr$-module structure you put on chain complexes.
If you do everything unbounded, it looks easier. Indeed, there is an algebra $\mathbb Z[d]/(d^2)$ in $Gr$ with $d$ in degree $-1$, and modules over that in $Gr$ are the same as chain complexes.
Then you have the following fact:

Let $C$ be a presentably symmetric monoidal category, and $A$ an algebra in $C$. Then monoidal structures on $\mathrm{LMod}_A(C)$ with a strong monoidal structure on the forgetful functor, and a "$C$-bilinear" structure on the tensor product correspond exactly to compatible coalgebra structures on $A$.

See my answer here, but this is exactly the same reason as why $E_n$-structures on $\mathrm{LMod}_A$ with $A$ as a unit correspond to $E_{n+1}$-structures on $A$.
I've stated it in a very imprecise way, and we'll say later that it will come back to bite us but let's pretend for a moment that it makes sense.
So now if you accept the extra adjectives that I've put on your tensor product (namely the extra requirement that it behaves nicely when you tensor with graded modules, i.e. chain complexes with $d=0$) you're reduced to classifying coalgebra structures on $\mathbb Z[d]/(d^2)$ in $Gr$, that is graded ring maps $\mathbb Z[d]/(d^2)\to \mathbb Z[d]/(d^2)\otimes \mathbb Z[d]/(d^2)$ satisfying the obvious requirements.
But now $d$ is in degree $-1$ and there isn't much in degree $-1$ in the codomain: $d$ must be sent to $n (d\otimes 1) + m (1\otimes d)$ for some $n,m\in\mathbb Z$.
Further, the square has to be $0$, but for that to make sense, you have to spot a hidden subtlety in what I said. I pretended that $Gr$ was symmetric monoidal, but there are several ways to do so. This impacts this squaring business, because it impacts how multiplication works in a tensor product of algebras: indeed, on $A\otimes B$ the product is defined by $A\otimes B \otimes A\otimes B \cong A\otimes A\otimes B\otimes B\to A\otimes B$.
If you plug in the algebraist's symmetric structure on $Gr$ (i.e. $x\otimes y \mapsto y\otimes x$), the squaring to $0$ tells you that $2nm (d\otimes d) = 0$, so $2nm = 0$, so $n=0$ or $m=0$. By symmetry we may assume $m=0$, so we just have $d\mapsto nd\otimes 1$. Coassociativity then tells me that $nd\otimes 1\otimes 1 = n^2 d\otimes 1\otimes 1$ so that $n^2 = n$, i.e. $n=1$ or $0$.
$n=0$ cannot happen because of co-unitality (I'll leave that as an exercise), so you're left with $n=1$. But for the same reason, $m$ cannot be $0$ by counitality. So there is no possibility here !
But now let's move to the more interesting question, of the topologist's/Koszul's symmetric structure on $Gr$ (i.e. $x\otimes y\mapsto (-1)^{|x||y|} y\otimes x$).
Then the squaring to $0$ condition looks very different ! Indeed, to multiply $nd\otimes 1 + m\otimes d$ with itself, you'll need to switch. Let's do it in steps: $(nd\otimes 1 + m\otimes d)^2 = (nd\otimes 1)^2 + (m\otimes d)^2 + (nd\otimes 1)(m\otimes d) + (m\otimes d)(nd\otimes 1)$. Because $|1| = 0$, the two first terms square to $0$ as expected. The third term multiplies to $mn d\otimes d$ because we're just switching $1$ and $m$ which are in degree $0$, and the fourth term however becomes $(-1)mn d\otimes d$. In other words, this automatically squares to $0$ !
So now let's look at co-unitality: plugging in $1\otimes d = 0$ should give the identity, so $n = m = 1$, and we get only one coalgebra map.
What's going on ??? You have given at least two monoidal structures (by the way, for your second one, I think the sign is on the wrong factor - think in terms of Koszul sign rules, as if $d$ was acting on the right), and I claim that there's only one.
Well first let's see what this coalgebra map actually gives you, as a differential. What's the action of $d$ on $M\otimes N$ ?
Well it's supposed to be given by $\mathbb Z[d]\otimes M\otimes N\to \mathbb Z[d]\otimes \mathbb Z[d]\otimes M\otimes N \to \mathbb Z[d]\otimes M \otimes \mathbb  Z[d]\otimes N$ so here you can clearly see what happens to $d\otimes m\otimes n$: it's mapped to $(d\otimes 1 + 1\otimes d)\otimes m \otimes n = d\otimes 1\otimes m \otimes n + 1\otimes d\otimes m \otimes n \mapsto d\otimes m\otimes 1\otimes n + (-1)^{|m|} 1\otimes m \otimes d\otimes n \mapsto d(m)\otimes n + (-1)^{|m|} m\otimes d(n)$, so you only get your first option.
Of course, your second option is also one (although I think you should put the sign on the other factor), so what went wrong here ? First of all, you can clearly see that this is an artifact of my encoding : had I encoded chain complexes as right modules, we would get the second option, so it must be something in my statement about the implicit words in my imprecise claim, namely what I really mean by "$C$-bilinear".
The point is that if I encode chain complexes as left modules, $\mathrm{LMod}_A(C)$ has a specified right $C$-action, so to compute the tensor product $\mathrm{LMod}_A(C)\otimes_C \mathrm{LMod}_A(C)$, I must do the usual trick of saying that a right module over a commutative ring has a canonical left action. But this trick uses of course the symmetry of $C$.
So even though $\mathrm{LMod}_A(C) \simeq \mathrm{RMod}_A(C)$ in this case, the module structures don't match up. Let's convince ourselves of that. Let $M$ be a chain complex and $C$ a graded abelian group. I want to tensor $M\otimes C$. In left modules, I have nothing more to say: the action is simply $\mathbb Z[d]\otimes M\otimes C\to M\otimes C$, so the differential is $\partial(m\otimes c) = \partial (m) \otimes c$.
If I instead view $M$ as a right module, and I tensor $M\otimes C$, the differential is twisted : $M\otimes C\otimes \mathbb Z[d]\to M\otimes \mathbb Z[d]\otimes C\to M\otimes C$, and so $\partial (m\otimes c) = (-1)^{|c|} \partial(m)\otimes c$.
So $\mathrm{LMod}_{\mathbb Z[d]}$ and $\mathrm{RMod}_{\mathbb Z[d]}$ are equivalent as categories, even as categories over $Gr$, but not as (say left) $Gr$-modules.
Of course, the classification theorem I mentioned works for a given $C$-module structure on $\mathrm{LMod}_A(C)$: the canonical one on the right !
So I guess at the end of the day I haven't answered your question... I've given an answer conditional to : fixing a $Gr$-module structure on chain complexes, and asking that chain complexes in fact be an algebra in $Gr$-modules (on top of the forgetful functor being stric monoidal). It follows that in these cases, there is only one monoidal structure that works, and it is in fact symmetric monoidal in the ways that we like (and for the same reason, if you ask for the forgetful functor to be symmetric monoidal and not just monoidal, it will be unique, because commutativity for algebra maps in $1$-categories is a property).
I have also not adressed the question of classifying $Gr$-module structures on chain complexes for which the forgetful functor is $Gr$-linear. It would be interesting to find out if there are others than the two I suggested. My guess would be no. By colimit-preservation blah, I assume it basically comes down to computing the differential on $M\otimes \mathbb Z[n]$ and by associativity of the module structure, on $M\otimes \mathbb Z[1]$. In other words, finding a self-equivalence of chain complexes which is the identity on underlying graded abelian groups.
By using the "stupid truncations", you reduce to finding a self equivalence of $Ab^{\Delta^1}$ which is the identity on $Ab^{\partial \Delta^1}$. And then by using representing objects such as $\mathbb Z\to \mathbb Z$ you see that it can only be $f\mapsto f$ or $f\mapsto -f$.
If you put that all back together, I think you get an argument that the only two $Gr$-module structures on chain complexes for which the forgetful functor is $Gr$-linear are the ones described earlier: I'll leave the details as an exercise :D
