= Solution
For the <free monoid> on $x,y$, use the function
$$
f(w,n)=\begin{cases}
1,&|w|>n\text{ and the final letter of }w\text{ is }x,\\
0,&\text{otherwise}.
\end{cases}
$$
The strict inequality is important. For any prefix $v$, $|vw|>|v|+n$ is equivalent to $|w|>n$. Whenever it holds, $w$ is nonempty and prefixing $v$ does not change its final letter. When it fails, both values are zero. Thus
$$
f(vw,n+|v|)=f(w,n),
$$
which proves equivariance for the diagonal action on $M\times\mathbb N$ and the trivial action on $B$. Hence $f$ is an element of the exponential described above. Let $g$ be the constant-zero equivariant function. They differ at $(x,0)$.
But every word $wy$ ends in $y$, so
$$
(y\cdot f)(w,n)=f(wy,n)=0=(y\cdot g)(w,n)
$$
for every $w,n$. The action of $y$ on $B^A$ is not injective, and \b[$B^A$ is not decidable], even though $B$ is decidable. This is a <nondecidable exponential of decidable monoid sets>. The <monoid> condition in part (a) also fails here: $|pmq|=|p|+|m|+|q|$ cannot equal $|p|$ for a nonempty word $m$.
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