I'm new to category theory, and I'm trying to define two categories which are not isomorphic, yet have identical graphs.
Category $A$ has 3 objects: {0, 1, 2}. For each pair of objects $m, n \in A$ there is a morphism $f_{m} : m \xrightarrow{f_{m}} n$ iff $(m + m) \% 3 = n$. Each object additionally has a morphism $id_{m} : m \xrightarrow{id_{m}} m$. Composition $f_{n} \circ f_{m} = id_{m}$ whenever the codomain of $f_{m}$ is equal to the domain of $f_{n}$.
The objects of category $B$ are 3 kinds of flowers: {"Bluebell", "Texas Bluebonnet", "Red Poppy"}. For each pair of (not necessarily distinct) flowers $p, r \in B$ there is a morphism $h_{pr} : p \xrightarrow{h_{pr}} r$ iff $p$ and $r$ have the same color petals (as implied by their names). The composition $h_{qr} \circ h_{pq}$ is equal to $h_{pr}$.
Functor $F : A \rightarrow B$ maps objects $F(2) = $"Bluebell", $F(1) =$ "Texas Bluebonnet", and $F(0) = $"Red Poppy". It also maps identities of $A$ according to $F(id_{m}) = h_{F(m)F(m)}$, where $m \in A$, and maps the modulo morphisms of $A$ according to $F(f_{m}) = h_{F(m)F(n)}$ where $n = (m + m) \% 3$.
Functor $G : B \rightarrow A$ maps objects $G($"Bluebell"$) = 2$, $G($"Texas Bluebonnet"$) = 1$, and $G($"Red Poppy"$) = 0$. It also maps morphisms of B according to $G(h_{pp}) = id_{G(p)}$, and $G(h_{pr}) = f_{G(p)}$, where $p,r \in B$ and $p \neq r$.
The categories are not isomorphic, because $F(f_{0}) = F(id_{0}) = h_{BB}$, while $G(h_{BB}) = id_{0}$ (abbreviating "Bluebell" as $_{B}$). But the graphs should still be the same, because identity morphisms do not appear in the graph of a category. Is this correct?