Let $G$ be a finite group. Usually, a 2-cocycle on $G$ with values in $\mathbb{Z}\_2 = \{+1, -1\}$ is a collection of signs $\epsilon_{g,h} \in \{+1, -1\}$, $g,h \in G$, satisfying the cocycle equation (written multiplicatively) $$ \epsilon_{g,h} \epsilon_{gh, k} = \epsilon_{h,k} \epsilon_{g,hk} $$ And a 2-coboundary is a 2-cocycle with $\epsilon_{g,h} = \frac{t_g t_g}{t_{gh}}$ for all $g,h \in G$, with $t : G \rightarrow \{+1, -1\}$ an arbitrary map. Then the second cohomology group is $H^2(G, \mathbb{Z}\_2)$ = {2-cocycles} / {2-coboundaries}.

But suppose we demand that our 2-cocyles satisfy the 2-cocycle equation above together with the "conjugate-cyclic" symmetry $$ \epsilon_{g,h} = \epsilon_{h^{-1} g^{-1}, g} $$ as well as the "conjugate symmetric" symmetry, $$ \epsilon_{g,h} = \epsilon_{h^{-1}, g^{-1}}. $$ These symmetries make sense from a TQFT perspective if you draw the 2-cocycle as a bunch of trivalent vertices, when the first symmetry corresponds to counterclockwise rotation of the diagram and the second to a kind of vertical flip.

And suppose now that the coboundaries given by $\{t_g\}$ satisfy $t_1 = 1$ and $t_g t_{g^{-1}} = 1$. This ensures that the "symmetric" 2nd cohomology group $H^2_{sym} (G) := ${ "symmetric" 2-cocycles} / {"symmetric" 2-coboundaries} makes sense.

Question: Does this symmetric 2nd cohomology group always vanish?

I've only checked one example, namely $G = \mathbb{Z}\_2 \times \mathbb{Z}\_2 = \langle a,b : a^2 = b^2 =(ab)^2 = 1 \rangle$. In normal cohomology, we have $H^2(G, \mathbb{Z}\_2)$ = $(\mathbb{Z}\_2)^3$ but according to my calculations, none of the non-trivial 2-cocycles respect the above symmetries, so $$ H^2_{sym} (G, \mathbb{Z}_2) = 0. $$ Is this perhaps always true for arbitrary $G$? Or perhaps I have made a silly mistake.