= Solution
Let $M\times X$ have the diagonal <right monoid action> $(g,x)m=(gm,xm)$, and put $E=\operatorname{Hom}_M(M\times X,Y)$. Define the right action on this set of <equivariant maps of monoid sets> by
$$
(ek)(g,x)=e(kg,x).
$$
It really stays in $E$: $(ek)(gm,xm)=e(kgm,xm)=e(kg,x)m$. Also $e1=e$ and $((ek)l)(g,x)=e(klg,x)=(e(kl))(g,x)$, so it satisfies the right-action law. Left multiplication in the first argument is intentional; no commutativity of $M$ has been assumed.
Define the <evaluation map of an exponential object>
$$
\operatorname{ev}:E\times X\to Y,\qquad \operatorname{ev}(e,x)=e(1,x).
$$
It is equivariant, because
$$
\operatorname{ev}(ek,xk)=e(k,xk)=e((1,x)k)=e(1,x)k.
$$
For any right $M$-set $Z$ and <equivariant map> $h:Z\times X\to Y$, define its <currying> by
$$
\widehat h(z)(g,x)=h(zg,x).
$$
For fixed $z$, this is equivariant in the diagonal variables: $h(zgm,xm)=h(zg,x)m$. The map $z\mapsto\widehat h(z)$ is also equivariant, since
$$
\widehat h(zk)(g,x)=h(zkg,x)=(\widehat h(z)k)(g,x).
$$
It satisfies $\operatorname{ev}(\widehat h(z),x)=h(z,x)$.
Conversely, given an equivariant $u:Z\to E$, set $h(z,x)=u(z)(1,x)$. Evaluation makes this equivariant, and currying recovers $u$:
$$
\widehat h(z)(g,x)=u(zg)(1,x)=(u(z)g)(1,x)=u(z)(g,x).
$$
These constructions are inverse and natural in $Z$. Therefore they establish the <exponential object> universal property
$$
\boxed{\operatorname{Hom}_M(Z,E)\cong\operatorname{Hom}_M(Z\times X,Y),\qquad
Y^X=E.}
$$
This is the <exponential of right monoid actions>, not the set of ordinary maps $X\to Y$ with an arbitrarily guessed action.
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