= Solution
A decidable $B$ has injective action maps. First prove that evaluation at $1$ distinguishes equivariant functions. Suppose $f(1,a)=g(1,a)$ for every $a$. Fix $m\in M$ and choose $p,q$ with $pmq=p$. For each $a$, equivariance gives
$$
p\cdot f(m,a)=f(pm,p\cdot a)
=f(pm,pm\cdot(q\cdot a))
=pm\cdot f(1,q\cdot a).
$$
The same equation holds for $g$, so these values agree. Cancel the injective action of $p$ on $B$ to obtain $f(m,a)=g(m,a)$. Hence $f=g$. This proves the hinted contrapositive and, more precisely, injectivity of the trace map $f\mapsto(a\mapsto f(1,a))$.
Now suppose $m\cdot f=m\cdot g$ in the exponential. Evaluating this equality at $(1,m\cdot a)$ gives
$$
f(m,m\cdot a)=g(m,m\cdot a),\qquad
m\cdot f(1,a)=m\cdot g(1,a).
$$
Cancel the action of $m$ on $B$. The trace maps agree, so the preceding argument gives $f=g$. Every action map on $B^A$ is therefore injective. By the introductory criterion, \b[$B^A$ is decidable whenever $B$ is] under the specified <monoid> condition. Neither cancellation in $M$ nor injectivity of its action on $A$ is assumed.
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