Suppose that $\phi$ is the abelianization of some $G$. Moreover, suppose that $\phi(x) = \phi(y)$ for some $x,y \in G$. Is it true that this occurs if and only if $y = gxg^{-1}$ for some $g \in G$?
The one direction is trivial: if $y = gxg^{-1}$, then $\phi (y) = \phi(gxg^{-1}) = \phi(g)\phi(x)\phi(g)^{-1} = \phi(x)$. But what about the other?
In fact what is true is that $\phi(x)=\phi(y) \iff xG'=yG'$. That is, $x$ and $y$ are representatives of the same coset of $G'$. So this is true iff $y^ {-1}x$ is a product of commutators.If for example $y=gxg^ {-1}$, then $y^ {-1}x=[g^ {-1},x]$. Hence conjugacy classes collapse to a singleton under $\phi$.
This equivalence only holds if $\phi$ is injective, in which case it's trivial. For if $\phi$ is not injective, then there is some $x\neq 1$ with $\phi(x)=\phi(1)=1$, and $x$ is not conjugate to $1$.
Let $G = S_5$, then $\phi:S_5 \rightarrow \{-1,1\}$ is the sign homomorphism. We have $\phi(\operatorname{id})$ = 1 = $\phi( (1,2) (3,4) )$, but obiviously $\operatorname{id}$ and $(1,2)(3,4)$ are not conjugates.