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 $\pi = F\_2$.) 



Are two elements of a group determined up to simultaneous conjugacy by the conjugacy classes of all of their products?
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.


