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Give $M\times X$ the diagonal right monoid action. On its equivariant maps into $Y$, define $(ek)(g,x)=e(kg,x)$. The evaluation map of an exponential object is $e(1,x)$ and currying of $h:Z\times X\to Y$ is $\hat{h}(z)(g,x)=h(zg,x)$. Both maps are equivariant, and they are inverse under evaluation. The left multiplication $kg$ in this right action is essential in a noncommutative monoid.

Let $M\times X$ have the diagonal right monoid action $(g,x)m=(gm,xm)$, and put $E={\mathrm{Hom}}_M(M\times X,Y)$. Define the right action on this set of equivariant maps of monoid sets byIt 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 objectIt is equivariant, becauseFor any right $M$-set $Z$ and equivariant map $h:Z\times X\to Y$, define its currying byFor fixed $z$, this is equivariant in the diagonal variables: $h(zgm,xm)=h(zg,x)m$. The map $z\mapsto \hat{h}(z)$ is also equivariant, sinceIt satisfies $\mathrm{ev}(\hat{h}(z),x)=h(z,x)$.

Conversely, given an equivariant $u:Z\to E$, set $h(z,x)=u(z)(1,x)$. Evaluation makes this equivariant, and currying recovers $u$:These constructions are inverse and natural in $Z$. Therefore they establish the exponential object universal propertyThis is the exponential of right monoid actions, not the set of ordinary maps $X\to Y$ with an arbitrarily guessed action.