I'm confused that is a function which is a bijective homomorphism enough to prove the two groups are isomorphic, or we need all possible maps to be bijective homomorphism?
I saw a statement that even though a function $\varphi$ is not an isomorphism, it cannot ensure that the two groups are not isomorphic because there may exist other isomorphisms which hold. Is that true?
I try my best to make my question clear but I've already got myself messy so please forgive and help. Thanks!
On
Two things being isomorphic means that there exists a homomorphism which is an isomorphism, not that all homomorphisms are isomorphisms. The latter would be quite a useless concept anyway, since between any two groups there is always the trivial homomorphism which sends everything to the neutral element. But this map is only an isomorphism if both groups contain only the neutral element. So only trivial groups could be isomorphic.
Yes, if $\varphi: G\to H$ is a bijective group homomorphism from a group $G$ to a group $H$, then $G\cong H$; its inverse $\varphi^{-1}:H\to G$ is also a bijective homomorphism.
It is not enough to conclude $A\not\cong B$ for groups $A,B$, from just $\psi:A\to B$ not being a bijective homomorphism. For example, for $S_3$ (the symmetric group of degree three) and $D_3$ (the dihedral group of order six), we have $S_3\cong D_3$ despite
$$\begin{align} \theta:S_3 &\to D_3,\\ x&\mapsto e \end{align}$$
not being a bijective homomorphism (although it is a homomorphism).