Need help discerning category isomorphism I'm new to category theory, and I'm trying to define two categories which are not isomorphic, yet have identical graphs. 


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*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?
 A: The definitions of the categories are fundamentally flawed. I suggest that you go back to the definition of a category and try to understand why what you've written down doesn't work out. A morphism is an arrow from one object to another, so "its only morphism is addition modulo $3$" doesn't make sense: addition modulo $3$ assigns a third number to two numbers, which isn't what we need to define a morphism. The same for path composition, which assigns a third path to two paths.
Here are three categories that have some similarity with what you've written; perhaps one or two of them are what you were trying to get at.
Category $A'$ has one object, the set $\mathbb Z_3=\{0,1,2\}$. For every $n\in\mathbb Z_3$, there is one morphism from $\mathbb Z_3$ to $\mathbb Z_3$, namely the function that sends $x$ to $x+n\bmod3$. Composing the morphisms corresponding to $n_1$ and $n_2$ yields the morphism corresponding to $n_1+n_2\bmod3$. The identity morphism for the only object is the one corresponding to $n=0$.
Category $A''$ has three objects, $0$, $1$ and $2$. There is one morphism for each pair of objects. Composition yields the only morphism between the appropriate objects, and the identity morphism for an object is the only morphism from that object to itself.
Category $B'$ has three objects, $a$, $b$ and $c$. There is one morphism from $x$ to $y$ for each path from $x$ to $y$ in the complete graph on $\{a,b,c\}$. Composition of morphisms is defined by composition of the corresponding paths. The identity morphism for an object is the path beginning and ending at that object without any edges.
Now we can define a functor $F$ from $B'$ to $A''$ that assigns to the objects $a$, $b$, $c$ the objects $0$, $1$, $2$, respectively, and to the morphism corresponding to a path from $x$ to $y$ the only morphism from $F(x)$ to $F(y)$. We can also define a functor $G$ from $A''$ to $A'$ that assigns to each object in $A''$ the only object in $A'$, and to the morphism from $x$ to $y$ the morphism that maps $x$ to $y$. We can also define the composition $H=G\circ F$ from $B'$ to $A'$.
