The set-up is that $G$ is a group, $C:=[G,G]$ the commutator subgroup with $G/C\cong \mathbb{Z}$. We assume that $x\in G$ is sent to a generator in $G/C$. We need to show that the map $t: C/[C,C] \to C/[C,C]$, $t[c]=[xcx^{-1}]$ is a well-defined autormorphism (here $[\cdot]$ denotes the coset in $C/[C,C]$).
I am not really sure how to get started on this problem. I am even somewhat confused by the notation. $x\in G$, so what do we even mean by $xcx^{-1}$. I would really appreciate a hint to help me get started. For context, this is part of a question in algebraic topology ($G=\pi_1 X$, $G/C\cong H_1(X)\cong \mathbb{Z}$).
I note that there is no $X$ in your set up. As for $x$, it is an element of $G$ that is sent to a generator of $G/C$.
I'm also assuming your commutator convention has $[x,c]=xcx^{-1}x^{-1}$ for the purposes of the answer below; if you use the other convention (with $[a,b]=a^{-1}b^{-1}ab$), then this requires a bit of a tweak.
Now, first note that if $c$ is an element of $[G,G]$ (that is, a product of commutators), then since $xcx^{-1} = xcx^{-1}c^{-1}c = [x,c]c\in [G,G]$. (Also, $[G,G]$ is fully invariant, so conjugation maps elements of $[G,G]$ to themselves).
So, suppose that $c\equiv c'\pmod{[C,C]}$. that means that there exists $u\in [C,C]$ such that $c=c'u$. Then $xcx^{-1}=x(c'u)x^{-1} = xc'x^{-1}(xux^{-1})$. But $[C,C]$ is also a fully invariant subgroup of $G$, hence invariant under conjugation. Therefore, $xux^{-1}\in [C,C]$, hence $xcx^{-1}\equiv xc'x^{-1}\pmod{[C,C]}$. Thus, $t$ is well-defined.
To verify it is an homomorphism, note that $t[cd] = [xcdx^{-1}]=[xcx^{-1}][xdx^{-1}]$.
To verify it is surjective, note that if $c\in C$, then $x^{-1}cx=x^{-1}cxc^{-1}c = [x^{-1},c]c\in C$; so $t[x^{-1}cx]=[c]$. Thus, $t$ is surjective.
Finally, if $t[c]=[e]$, then $xcx^{-1}\in [C,C]$. Therefore, $c\in x^{-1}[C,C]x = [C,C]$, so $[c]=[e]$.