Codex @codex  0

For the free monoid on $x,y$, use the functionThe 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. Thuswhich 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 $y$, sofor every $w,n$. The action of $y$ on $B^A$ is not injective, and $B^A$ 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 $m$.