Are two elements of a group determined up to simultaneous conjugacy by the conjugacy classes of all of their products? Let $G$ be a group (if it helps, assume that $G$ is a Lie group or finite). Is a pair of elements $(g, h) \in G \times G$ determined up to simultaneous conjugacy by the conjugacy class of every element $w(g, h) \in G$, where $w$ runs over all words in the free group on two generators? 
If $G$ is finite, can we bound the length of the words $w$ needed in terms of $|G|$? 
If the answer to the above question is positive, let $\pi$ be a second group (if it helps, assume that $\pi$ is finitely presented). $G$ acts on the set $\text{Hom}(\pi, G)$ by pointwise conjugation. Is an element $\phi \in \text{Hom}(\pi, G)$ determined up to conjugacy by the conjugacy class of every element $\phi(w)$ where $w \in \pi$? (The above is the special case $\pi = F_2$.) 
 A: Just saw this old question thanks to the link from
the new 
MO question #212929.  Another flavor of counterexample:
let $G$ be the $ax+b$ group over some field $k$, and let
$t(b) \in G$ be the transformation $x \mapsto x + b$.
Then the $t(b)$ with $b \neq 0$ are all conjugate, but
a pair $(t(b_1),t(b_2))$ is simultaneously conjugate with
$(t(b'_1),t(b'_2))$ iff $b_1/b_2 = b'_1/b'_2$.
Suppose, then, that $b_1/b_2 \neq b'_1/b'_2$, 
and that both quotients are irrational
(i.e. if $mb_1=nb_2$ or $mb'_1=nb'_2$ for some integers $m,n$ 
then $m=n=0$ in $k$).  Then the pairs
$(g,h) := (t(b_1),t(b_2))$ and $(g',h') := (t(b'_1),t(b'_2))$
are not related by simultaneous conjugacy, but cannot be
distinguished by the conjugacy class of any $w(g,h)$.
The same is true if we take $G = {\rm SL}_2(k)$
or $G = {\rm PSL}_2(k)$ and work in its $ax+b$ subgroup
$({* \; * \atop 0 \; *})$ [with $t(b) = ({1 \; b \atop 0 \; 1})$].
Indeed that was the example that first came to mind, 
suggested by the restriction to diagonalizable elements in
MO 212929
(NB the $t(b) \in {\rm SL}_2(k)$ with $b\neq 0$ are not diagonalizable).
In the smallest finite examples of this kind, $k$ is the 4-element field, and
$b_1/b_2$, $b'_1/b'_2$ are the two elements of $k$ other than $0$ and $1$.
Then the $ax+b$ group is isomorphic with $A_4$, and 
${\rm SL}_2(k) \cong {\rm PSL}_2(k) \cong A_5$,
and in each case we can take
$g=g'=(12)(34)$, $h=(13)(24)$, and $h'=gh=(14)(23)$. 
[The pairs $(g,h)$ and $(g',h')$ become conjugate in $S_4$ and $S_5$,
which are the extensions of the $ax+b$ group and ${\rm SL}_2(k)$
by the Galois automorphism of $k$ that also switches the two
irrational elements of $k$.]
A: Suppose a group $G$ had the property that n-tuples $\lbrace x_1,\dots, x_n\rbrace$ and $\lbrace y_1,\dots,y_n\rbrace$ satisfy: if $w(x_1,\dots,x_n)$ is conjugate to $w(y_1,\dots,y_n)$ for all $w\in F_n$ then there is a uniform conjugator $g$ so that $y_i=gx_ig^{-1}$. Then as a corollary you get that if an endomorphism of $G$, satisfies $\varphi(x)$ is conjugate to $x$ for all $x$ then $\varphi$ is an inner automorphism.
However there are non-examples to this property in several classes of groups, including finite groups. The original property does however hold for torsion-free $\delta$-hyperbolic groups. This is proved in the paper "On endomorphisms of torsion-free hyperbolic groups", which also has references to the previous work.
A: For a concrete example consider the symmetric group on 6 symbols. The pairs ((1,2)(3,4),(1,3)(2,4)) and ((1,2)(3,4),(3,4)(5,6)) both generate Klein 4-groups.
Words in them are either trivial of conjugate to (1,2)(3,4).
The generated subgroups (and thus the pairs) are not conjugate as one fixes two points.
