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.

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