Is the number of homomorphisms between two isomorphic groups equals to the number of Automorphisms for each group? Is the number of homomorphisms between two isomorphic groups equals to the number of Automorphisms for each group?
Let us first divide to cases:

*

*finite groups


*infinite groups
Let $G,H$ be some groups such that: $G \cong H$.
Is the number of homomorphisms between $G\to H$ equals to the number of elements in Aut$(G)$?
My intuition is that the answer is true, at least for one of the cases, but I don't know how to prove that.
 A: It is false.
When $G=H=\mathbb{Z}$,
$\#$Hom$(G,H)=\#$Hom$(\mathbb{Z}, \mathbb{Z})=\# \mathbb{Z}=\infty.$
But $\#$Aut$(G)=\#$Aut$(\mathbb{Z})=\# \{\pm1\}=2$.
A: Fix an isomorphism $\varphi:G→ H$. Every isomorphism $\psi:G→ H$ can be seen as the composition of $\varphi$ with an automorphism of $H$. So there is a bijection between the set of isomorphism from $G$ to $H$, and the set of automorphisms of $H$.
EDIT: I read in a comment that you wrote homomorphisms. If you meant isomorphism, then my other answer is right. If you really meant homomorphism, the result you are talking about is clearly false, even in the case that $G=H$! In that case, every automorphism of $G$ is also a morphism from $G$ to $H$, but there are some morphisms from $G$ to $H$ (for example, the trivial morphism if $|G|\ge2$) which aren't automorphisms
A: To be explicit:
Let $I(G,H)$ be the set of Isomorphisms between $G$ and $H$. Take an element $\phi \in I(G,H)$:
Now $\phi^{-1} I(G,H) \subseteq Aut(G)$.
Now $\phi Aut(G)  \subseteq I(G,H)$. 
So composing $|I(G,H)| = |Aut(G)|$.
So the number of Isomorphisms is equal to the number of Automorphisms.
A: A finite group example (inspired by answer by Nakayama):
Let $G$ be a cyclic group of order $n$ generated by $g$.
Then $Hom(G,G) = \{f : f(g) = g^i, i \in [n-1] \}$.
Then $Aut(G,G) = \{f : f(g) = g^i, i \in [n-1], gcd(i,n) = 1 \}$.
Clearly $|Hom(G,G)| > |Aut(G)|$ for $n$ not a prime number.
