Is Yetter's invariant multiplicative under connected sum? Classical formulation
Consider the (untwisted) Dijkgraaf-Witten invariant, defined for an oriented, connected, closed manifold $M$ and a finite group $G$:
$$DW_G(M) := \lvert \operatorname{Hom}(\pi_1(M), G )\rvert $$
We're counting group homomorphisms from the fundamental group of the manifold to $G$. I am multiplying the more common definition by the size of the group, in order to simplify the following formulas. (Some are of the opinion that each homomorphism should be weighted by the inverse size of its automorphism group. I think this is not so, see e.g. this article. Instead, I think you can also sum over equivalence classes of $G$-bundles, which correspond to conjugacy classes of homeomorphisms, and there you need the weights.)
Now this invariant satisfies a simple identity, where $\#$ denotes connected sum:
$$DW_G(M_1 \# M_2) = \lvert \operatorname{Hom}(\pi_1(M_1 \# M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1) * \pi_1(M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1)) \rvert \cdot \lvert \operatorname{Hom} (\pi_1(M_2), G )\rvert = DW_G(M_1) \cdot DW_G(M_2) $$
The Dijkgraaf-Witten invariant is multiplicative under direct sum.
We've used the free product of groups, denoted as $*$. We know from the Seifert-van-Kampen theorem that $\pi_1(M_1 \# M_2) = \pi_1(M_1) * \pi_1(M_2)$.
Homotopy theory formulation
We can also write the Dijkgraaf-Witten invariant as the number of homotopy classes into the classifying space of $G$:
$$DW_G(M) = \lvert [M, BG] \rvert$$
Indeed, instead of $BG$ we could insert any (sufficiently finite) homotopy 1-type. This calls for a generalisation, known as the Yetter invariant.
An aside: The twisted invariant
Let $[M]$ be the fundamental class of the oriented manifold $M$, $\omega \in H^{\operatorname{dim} M}(G, k^\times)$ a group cohomology element (with values in the unit group of a field) and $c\colon M \to B(\pi_1(M))$ the canonical map. Then the twisted Dijkgraaf-Witten invariant is defined as:
$$DW_G^\omega(M) := \sum_{\phi\colon \pi_1(M) \to G} \langle \phi^*(\omega), c_*([M]) \rangle $$
The inner product $\langle - , - \rangle$ comes from Poincaré duality.
It is easy to see that if $\omega$ is the trivial cocycle, the sum ranges over $1$, and we recover the original formula.
Yetter invariant
David Yetter, and then later Tim Porter, defined, for a homotopy 2-type $\mathcal{T}$, the following invariant:
$$Y_\mathcal{T}(M) = \lvert [M, \mathcal{T}] \rvert$$
I think the original definition is more hands-on, in terms of crossed modules.
Edit: I seem to have omitted the correct groupoid cardinalities here. See Arun Debray's answer further below for the correct version, or this article.
Since there are some kinds of higher formulations of Seifert-van-Kampen (as discussed in this question) suitable for this situation, I'm asking the following question:
Question: Is the Yetter invariant multiplicative under connected sum? I.e. does the following hold:
$$Y_\mathcal{T}(M_1 \# M_2) = Y_\mathcal{T}(M_1) \cdot Y_\mathcal{T}(M_2)$$
 A: As Kevin Walker pointed out in a comment, Dijkgraaf-Witten invariants are weighted by $1/\mathrm{Stab}(\rho)$. In
the same way, the Yetter invariant for $\mathcal T$ and $M$ is generally defined such that it's weighted using the
2-groupoid cardinality of $\pi_{\le 2}\mathrm{Map}(M, \mathcal T)$, so that the invariant is
$$\sum_{[f\colon M\to\mathcal T]} \frac{|\pi_2(\mathrm{Map}(M, \mathcal T), f)|}{|\pi_1(\mathrm{Map}(M, \mathcal T),
f)|}.$$
If we use this normalization, the Yetter invariants are the partition functions of a TQFT $Z_{\mathcal T}$, usually
called the Yetter model. In this case, a different MathOverflow
answer by Kevin Walker tells us that $Z_{\mathcal T}$ is multiplicative
under connect sum iff


*

*$\dim Z_{\mathcal T}(S^{n-1}) = 1$, and

*$Z_{\mathcal T}(S^n) = 1$.


The state space $Z_{\mathcal T}(M^{n-1}) := \mathbb C[[M, \mathcal T]]$, so for $n = 3$, the first property doesn't
hold: $\dim Z_{\mathcal T}(S^2) = |\pi_2(\mathcal T)|$. A similar problem occurs for $n = 2$.
If $n > 3$, then $[S^{n-1}, \mathcal T] = 0$, so the first property holds. The second property does not quite hold:
$[S^n, \mathcal T] = 0$, but we have to calculate the weighting. Since $\mathrm{Map}(S^n, \mathcal
T)\simeq\mathrm{Map}(\mathrm{pt}, \mathcal T)\cong \mathcal T$,
$$Z_{\mathcal T}(S^n) = \frac{|\pi_2(\mathcal T)|}{|\pi_1(\mathcal T)|},$$
which is frequently not equal to 1.
I don't know about the unweighted version you mentioned, since it doesn't come from a TQFT as far as I know.
