The skew right unitor of the left tensor product of pointed sets is the natural transformation
at $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ is given by the composition
where $X\to X\vee X$ is the map sending $X$ to the second factor of $X$ in $X\vee X$.
(Proven below in a bit.)
In other words, $\smash {\rho ^{\mathsf{Sets}_{*},\lhd }_{X}}$ acts on elements as
i.e. by
for each $x\in X$.
The morphism $\smash {\rho ^{\mathsf{Sets}_{*},\lhd }_{X}}$ is non-invertible, as it is non-surjective when viewed as a map of sets, since the elements $x\lhd 0$ of $X\lhd S^{0}$ with $x\neq x_{0}$ are outside the image of $\rho ^{\mathsf{Sets}_{*},\lhd }_{X}$, which sends $x$ to $x\lhd 1$.
Firstly, note that, given $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, the map
is indeed a morphism of pointed sets as we have
Next, we claim that $\rho ^{\mathsf{Sets}_{*},\lhd }$ is a natural transformation. We need to show that, given a morphism of pointed sets
the diagram
