I am trying to find all natural transformations from the functor $\operatorname{id}_{\textsf{Grp}}$ to itself. $\require{AMScd}$ \begin{CD} G @>{\phi}>> H \newline @V{\alpha_G}VV @VV{\alpha_H}V \newline G @>{\phi}>> H. \end{CD}
That is, I am looking for a family ${(\alpha_G:G \to G)}_{G \in \textsf{Grp}_0}$ of morphisms such that the diagram above commutes. I figured out that defining for each $G$, $\alpha_G(x) = x$ or $\alpha_G(x) = 0$ would work respectively, and I felt like that would be all the options, but then I found this post and now I am not sure about my assertion anymore.
I tried to come up with more examples for defining $\alpha$, but without any success. For any two groups $G$ and $H$, I would want $\alpha_G$ and $\alpha_H$ to 'behave similarly', e.g. in the case of abelian groups $\alpha_G(x) = -x$ would work (since inverses are preserved by group homomorphisms).
However, $\alpha_G(x) = -x$ would only be an automorphism if $G$ is abelian, so I considered defining $\alpha_G$ such that it would send $x$ to its inverse if $G$ is abelian and to itself otherwise - just to come up with some example. Obviously this did not work, since for some abelian group $G$ and some non-abelian group $H$, we would have $\phi(x) = \alpha_H(\phi(x) = \phi(\alpha_G(x)) = \phi(-x)$.
Maybe my first intuition was right anyway. To prove that there are no other natural transformations from $\operatorname{id}_{\textsf{Grp}}$ to itself other than the identity transformation and the one sending everything to zero I would need to show that: If for any group $G$ there exists an automorphism $\alpha_G$ different to the trivial automorphism and the identity, then there exists a homomorphism $\phi$ from $G$ to some group $H$ such that we cannot find an automorphism $\alpha_H:H \to H$ such that $\phi \circ \alpha_G \neq \alpha_H \circ \phi$. But how would I go about this?
Let us write $n=\alpha_{\mathbb{Z}}(1)\in \mathbb{Z}$.
Now let $H$ be any group, and $x\in H$. Then there is a unique group morphism $\phi_x: \mathbb{Z}\to H$ such that $\phi_x(1)=x$. Apply the commutative diagram to show that $\alpha_H(x)=x^n$.
This means that $\alpha$ depends only on $n$. Now what are the values of $n\in \mathbb{Z}$ such that $x\mapsto x^n$ is a well-defined group morphism for all groups?