Tannaka Duality I only know two theorems for neutral Tannaka categories. 
(1) One states that the set of $k$-group scheme homomorphisms $m:G_1\to G_2$ is in one to one correspondence with the set of $k$-linear 
tensor functors $m^*:{\rm Rep}_k(G_2)\to {\rm Rep}_k(G_1)$ which commute with the forgetful fiber functors. 
(2) The other states that for any neutral Tannaka category $(C,\omega)$ there is a canonical $k$-linear tensor equivalence $\alpha_C:C\to {\rm Rep}_k(G_C)$ for some affine group scheme $G_C$, and the equivalence commutes with the fiber functors.
Question: Is it true that if I have a $k$-linear tensor functor $b:C\to C'$
 which commutes with the fiber functors, then there is a unique $k$-group scheme homomorphism $m_b:G_{C'}\to G_C$ which satisfies $\alpha_{C'}b=m_b^*\alpha_C$. Here "=" means strictly equal not an isomorphism of functors.
My problem is that (1) is a correspondence between group homomorphism and "strict"functors. There are different morphisms $m_1,m_2$ which induce two isomorphic functors i.e. $m_1^*\cong m_2^*$. For example, if $G$ is an affine $k$-group scheme which admits a non-trivial $k$-rational point $g\in G(k)$. Let $m_1: G\to G$ be the identity and $m_2: G\to G$ be the conjugation: $a\to gag^{-1}$. Then for any $V\in{\rm Rep}_k(G)$ we define an isomorphism $m_1^*(V)\to m_2^*(V)$ by the isomorphism $V\xrightarrow{ g\cdot}V$.
The reason I ask this is that if I have functors between neutral Tannakian categories $a_{12}:(C_1,\omega_1)\to (C_2,\omega_2)$, $a_{23}:(C_2,\omega_2)\to (C_3,\omega_3)$, and $a_{13}:(C_1,\omega_1)\to (C_3,\omega_3)$, and if we know $a_{13}=a_{23}a_{12}$, then I want to have also a commutative diagram of the corresponding Tannakian group schemes. If we only have $a_{13}\cong a_{23}a_{12}$ this is already false becasuse of the example I provided, but I would hope this could be true when we have a strict equality.
 A: I think you're confused. The issue in (1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between  a functor and itself. In that case you can choose the identity, but that won't have meaning here.
A commutative triangle of functors should be a seen as  trio $f,g,h$, with an isomorphism $g \circ f = h$.  This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.
In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)
