The join $\mathcal{C} \star \mathcal{C'}$ of two categories $\mathcal{C}, \mathcal{C'}$ is defined here: https://ncatlab.org/nlab/show/join+of+categories
A key feature is that if $X$ (resp. $Y$) are objects of $\mathcal{C}$ (resp. $\mathcal{C'}$) then there is exactly one morphism $X \rightarrow Y$ in the join, and no morphism $Y \rightarrow X$.
On the other hand, the join $\mathcal{S} \star \mathcal{S'}$ of two simplicial sets $\mathcal{S}, \mathcal{S'}$ is defined here: https://ncatlab.org/nlab/show/join+of+simplicial+sets#concrete_formulas
In the context of higher category theory, one wants to think of an edge $\phi:\Delta^1 \rightarrow \mathcal{S}$ in a simplicial set $\mathcal{S}$ as a morphism from the vertex $d_0(\phi)$ to the vertex $d_1(\phi)$.
But according to the formulas given in the second link above, the edges ($1$-simplices) in $\mathcal{S} \star \mathcal{S'}$ are $$(\mathcal{S} \star \mathcal{S'})_1 = \mathcal{S}_1 \cup \mathcal{S'}_1 \cup \mathcal{S}_0 \times\mathcal{S'}_0$$
and in particular, the face maps are defined on a new "mixed" edge $(X,Y) \in \mathcal{S}_0 \times\mathcal{S'}_0$ by: $$d_0(X,Y)=Y \in \mathcal{S'}_0 \subset (\mathcal{S} \star \mathcal{S'})_0$$ $$d_1(X,Y)=X \in \mathcal{S}_0 \subset (\mathcal{S} \star \mathcal{S'})_0$$ That is to say, we have a "morphism" $(X,Y)$ from $Y$ to $X$ in the join but not conversely. This is exactly opposite to the situation for categories.
Have I made a mistake somewhere, or are these operations truly incompatible with one another?