For a Grothendieck topos $\mathcal{E}$, are the following assertions equivalent?

$(i)$ $\mathcal{E}$ is localic.

$(ii)$ The diagonal geometric morphism $\mathcal{E} \to \mathcal{E} \times \mathcal{E}$ is an embedding. (Here $\mathcal{E} \times \mathcal{E}$ is the product topos, not the product category.)

$(iii)$ For every Grothendieck topos $\mathcal{E}'$, $\mathrm{Geom}(\mathcal{E}', \mathcal{E})$ is a preorder (no parallel geometric transformations).

The implications $(i) \Rightarrow (ii)$ and $(ii) \Rightarrow (iii)$ do hold:

- $(i) \Rightarrow (ii)$: Any diagonal morphism $X \to X \times X$ (in any category) is a split mono and a split mono of locales is an embedding. The (forgetful) functor from locales to toposes preserves the product and turns embeddings of locales into geometric embeddings.
- $(ii) \Rightarrow (iii)$: If $\mathcal{E} \to \mathcal{E} \times \mathcal{E}$ is an embedding, then the diagonal functor $\mathrm{Geom}(\mathcal{E}', \mathcal{E}) \to \mathrm{Geom}(\mathcal{E}', \mathcal{E} \times \mathcal{E}) \simeq \mathrm{Geom}(\mathcal{E}', \mathcal{E}) \times \mathrm{Geom}(\mathcal{E}', \mathcal{E})$ must be fully faithful. But this means precisely that $\mathrm{Geom}(\mathcal{E}', \mathcal{E})$ is a preorder.

So in summary, is a topos with only a preorder of $\mathcal{E}'$-based points for every $\mathcal{E}'$ already localic?