Amount of Isomorphisms Between Two Cyclic Groups of Identical Order is equal to the amount of Automorphisms of one of the cyclic groups This fact seems rather obvious but for some reason I can’t wrap my head around a proof.
Also, just generally, if we have two groups $G_1$ and $G_2$ equal up to isomorphism, then is the amount of automorphisms between the two equal to $|{\rm Aut}(G_1)|= |{\rm Aut}(G_2)|$?
Is there a proof for this? I have been trying but since I’m new to all this it feels like Im just getting more and more confused on what an isomorphism even is.
Thanks!
 A: Hint
There is only one cyclic group of given order.  Sometimes denoted $C_n$.  It is also $\Bbb Z_n$.
As to the other question,  isomorphic groups have isomorphic automorphism groups.   Let $\phi: G\to H$ be an isomorphism.   Then $\hat\phi:\rm {Aut} G\to\rm{Aut}H$ given by $\hat\phi(\psi)=\phi\circ\psi\circ \phi^{-1}$ is an isomorphism.
A: Here is more detailed answer. The statement doesn't depend on the groups being cyclic.

Let $\psi \in \text{Iso}(G_1, G_2)$. Then for each $\phi \in \text{Aut}(G_2)$ we can construct an isomorphism as follows:
$$ g \mapsto \phi(\psi(g)).$$
Thus, we have an injective map $\text{Aut}(G_2) \to \text{Iso}(G_1,G_2)$.
Let $\hat{\psi}$ be an isomorphism from $G_1$ to $G_2$. Now the map
$$ g \mapsto \hat{\psi}(\psi^{-1}(g)) $$
is an automorphism on $G_2$. Thus, choosing $\phi$ to be this automorphism, we see that the map is also surjective.
Thus the two have the same cardinality.

Note this had nothing to do with them being cyclic groups. Philosophically, two groups being isomorphic mean that they are equal. Thus an isomorphism is in some an automorphism, (since the two groups are "equal" and vice verse).
