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?
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.
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).
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.
