Context
Let $A=\bigoplus_{i,k} A^i_k$ be a bigraded module and let us say we have a linear map $\circ :A\otimes A\to A$ of bidegree $(0,0)$. This operation satisfies certain associativity relation with respect to the braiding morphism $\gamma(a\otimes b)=(-1)^{ij+kl}b\otimes a$ for $a\in A^i_k$ and $b\in A^j_l$ (we say that $a$ has bidegree $(i,k)$ and $b$ has bidegree $(j,l)$).
A morphism $f:A\to B$ that respects this structure is a linear map $f:A\to B$ of bidegree $(p,q)$ such that $f(a\circ b)=(-1)^{pi+qk}f(a)\circ f(b)$.
Problem
If I modify this operation by $a\bar{\circ}b=(-1)^{l(i+k)}a\circ b$, I get the same associativity relation with respect to the braiding $\gamma'(a\otimes b)=(-1)^{(i+k)(j+l)}b\otimes a$. I would like to define $\bar{f}=\pm f$ so that $\bar{f}(a\bar{\circ}b)=(-1)^{(p+q)(i+k)}\bar{f}(a)\bar{\circ}\bar{f}(b)$, in other words $\bar{f}$ respects $\bar{\circ}$ with respect to the braiding $\gamma'$. The sign factor of $\bar{f}$ should depend on the bidegree components of $f$ and its input.
Attemp
Let us say $f(a)=(-1)^{\varepsilon(a)}f(a)$. Now we have
$$\bar{f}(a\bar{\circ} b)=(-1)^{l(i+k)}\bar{f}(a\circ b)=(-1)^{l(i+k)+\varepsilon(a\circ b)}f(a\circ b).$$
Using that $f$ respects $\circ$ with respect to $\gamma$ we have
$$\bar{f}(a\bar{\circ} b)=(-1)^{l(i+k)+\varepsilon(a\circ b)+pi+qk}f(a)\circ f(b).$$
Since $f$ has bidegree $(p,q)$, $f(a)$ has bidegree $(i+p,k+q)$ and $f(b)$ has bidegree $(j+p,l+q)$ so now we have
$$\bar{f}(a\bar{\circ} b)=(-1)^{l(i+k)+\varepsilon(a\circ b)+pi+qk+(l+q)(i+p+k+q)}f(a)\bar{\circ} f(b).$$
Finally, we have
$$\bar{f}(a\bar{\circ} b)=(-1)^{l(i+k)+\varepsilon(a\circ b)+pi+qk+(l+q)(i+p+k+q)+\varepsilon(a)+\varepsilon(b)}\bar{f}(a)\bar{\circ} \bar{f}(b).$$
And therefore that huge exponent should have the same parity as $(p+q)(i+k)=pi+pk+qi+qk$. However, cancelling some equal terms this implies
$$\varepsilon(a\circ b)+\varepsilon(a)+\varepsilon(b)+l(p+q)+q(p+k+q)\equiv pk.$$
Any reasonable choice of $\varepsilon(a)$ implies $\varepsilon(a\circ b)=\varepsilon(a)+\varepsilon(b)$ since the bidegree of $a\circ b$ is $(i+j,k+l)$, so all the $\varepsilon$'s would vanish and the resulting equation modulo 2 would not be true in general.
Am I making any mistake or it is not possible to define such $\bar{f}$?
A map preserving $\circ$ must be of bidegree $(0,0)$. Indeed, $f$ of biegree $(p,q)$ must satisfy
$$f(a\circ a)=\pm f(a)\circ f(a).$$
for $a$ of bidegree $(i,k)$. The left hand side has bidegree $(2i+p,2k+q)$, while the right hand side has bidegree $(2i+2p,2k+2q)$, which implies $2p=p$ and $2q=q$, so $p=q=0$ (if $2\neq 0$ and we are grading over domains, which I was actuallly assuming). Therefore, $f$ automatically preserves $\bar{\circ}$.
Edit: if $2=0$, then $2p=p$ also implies $p=0$ and similarly with $q$. If $2\neq 0$ and the degrees are not elements of a ring with the cancellation property I don't know what happens.