Can we classify all finite 2-generated groups $G$ such that if $x,y$ generate $G$, then so does $x,yxy^{-1}$? Can we classify finite 2-generated groups $G$ satisfying the following property:
For any pair $x,y$ which generate $G$, the pair $x,yxy^{-1}$ also generates $G$.
By the comments, no nontrivial abelian group can satisfy this property, so I suppose the first question is: Do such groups $G$ exist?
 A: This is only a long comment.


Claim. If $G$ is a $2$-generated finite group such that $(x, yxy^{-1})$ generates $G$ whenever $(x, y)$ does, then $G$ is perfect.
Proof. Following this MSE answer, we can assume that the abelianization $G_{ab} \Doteq G/[G, G]$ of $G$ is cyclic. As a result, the group $G$ has a generating pair a component of which lies in $[G, G]$. To see this, apply Nielsen transformations to the image of a generating pair of $G$ in $G_{ab}$. Hence $G = [G, G]$.


As a perfect soluble group is just a trivial group in disguise, there is no non-trivial finite soluble group satisfying OP's condition (which echoes a discussion initiated in the comments).
Side note. Since OP's property is stable under taking quotients (use Gaschutz's Lemma to lift generating pairs), the non-existence of such groups will be established if it is shown that no finite simple groups satisfy OP's property (see YCor's comment for early work in this direction).
A: Theorem The only finite group satisfying the condition is the trivial group.
Proof. Let $G$ be a nontrivial finite group and $S$ be a simple quotient of $G$, which satisfies the condition by Gaschutz's lemma. Then $S$ is non-abelian as mentioned above. If $x\in S$ is any involution, then by  well-known result of Guralnick and Kantor in Probalistic generation of finite simple groups, there exists an element $y\in S$ such that $S=\langle x,y\rangle$ while $\langle x,yxy^{-1}\rangle$ is a dihedral group, that is, $S\neq\langle x,yxy^{-1}\rangle$, a contradiction.
A: As shown in this MSE answer, the subgroup generated by a conjugacy class in any finite group with a non-cyclic abelianization is proper. So, any such 2-generated group is a counterexample.
