If $x\in G$ commutes with its conjugates by $y,z\in G$, does it commute with its conjugates by $yz$ and $y^{-1}$? In a group $G$, let $[x,y]=x^{-1}y^{-1}xy$ be the commutator of $x$ and $y$. Suppose that $x$, $y$ and $z$ are elements in $G$ satisfying 
$[x,y^{-1}xy]=[x,z^{-1}xz]=1$. 
Is it necessarily true that $[x,(yz)^{-1}x(yz)]=1$?
Is it necessarily true that $[x,yxy^{-1}]=1$?
 A: The first statement in general is false; Tobias Kildetoft gives a (bunch of) counterexample(s).
The second statement can be verified by computation:
\begin{eqnarray*}
[x,y^{-1}xy]=1 &\Leftrightarrow&\ (x)^{-1}(y^{-1}xy)^{-1}(x)(y^{-1}xy)=1\\
&\Leftrightarrow&\ (x^{-1})(y^{-1}x^{-1}y)(x)(y^{-1}xy)=1\\
&\Leftrightarrow&\ y(x^{-1})(y^{-1}x^{-1}y)(x)(y^{-1}xy)y^{-1}=1\\
&\Leftrightarrow&\ (yx^{-1}y^{-1})(x^{-1})(yxy^{-1})(x)=1\\
&\Leftrightarrow&\ (yxy^{-1})^{-1}(x)^{-1}(yxy^{-1})(x)=1\\
&\Leftrightarrow&\ [yxy^{-1},x]=1\\
&\Leftrightarrow&\ [x,yxy^{-1}]=1.
\end{eqnarray*}
An easier way to see this is to note that $[a,b]=1$ if and only if $ab=ba$. Then
$$[x,y^{-1}xy]=1\ \Leftrightarrow\ xy^{-1}xy=y^{-1}xyx\ \Leftrightarrow\ yxy^{-1}x=xyxy^{-1}\ \Leftrightarrow\ [x,yxy^{-1}]=1.$$
A: Denote $x^g = g^{-1}xg$. As mentioned in the comments, using the identity $[a,b]^g = [a^g, b^g]$ allows us to deduce that the second statement is true. That is, $[x, x^y] = 1$ implies $[x, x^{y^{-1}}] = 1$.
Assume that $[x, x^y] = [x, x^z] = 1$. The first statement asks whether this implies $[x, x^{yz}] = 1$. Now $[x, x^{z^{-1}}] = 1$, so if the first statement were true, we would have $[x, x^{yz^{-1}}] = 1$, which is equivalent to $[x^z, x^y] = 1$.
Thus if $[x, x^y] = [x, x^z] = 1$ and $[x^z, x^y] \neq 1$, we get the counterexample $[x, x^y] = [x, x^{z^{-1}}] = 1$, $[x, x^{yz^{-1}}] \neq 1$. In other words, to find a counterexample to the first statement, it suffices to find two conjugates of $x$ that commute with $x$, but do not commute with each other. Intuitively it seems that this should be possible. Just because two elements commute with $x$, why should they commute with each other?
Finding counterexamples in symmetric groups is pretty straightforward. Two elements are conjugate if and only if they have the same cycle type. Also, disjoint permutations commute with each other. With this and the previous remark, you can figure out as many examples as you want similar to the one given by Tobias in the comments. For example consider $x = (12)$, $x^y = (34)$, $x^z = (45)$.
