Let $G$ be a group generated by a conjugacy class $C$. I am interested in studying this property:
for every $x,y\in C$ there exists $h\in \langle x,y\rangle$ such that $y=hxh^{-1}$.
Basically the property says that the conjugacy class $C$ does not split when you consider subgroups of $G$. I am wondering if this property (which to me sounds quite strong) forces some property on the group $G$, in particular in terms of solvability.
Let me name a random non-soluble example for you. Let $H$ be the Suzuki simple group $Sz(8)$ and let $G=\operatorname{\rm Aut}(H)$, so that $|G:H|=3$. Crucially, $|H|$ is not divisible by $3$. Let $x$ be an element of order $3$ in $G$, and let $C$ be the conjugacy class of $x$. If $y$ is any conjugate of $x$, the elements $x$ and $y$ are conjugate in $\langle x,y\rangle$ by conjugacy of Sylow $3$-subgroups.
The only finite simple groups with outer automorphisms of coprime order are the groups $Sz(2^{2n+1})$, but you can make infinitely many examples of this kind. (Edit: Geoff Robinson points out that there are other finite simple groups with this property, such as $\operatorname{\rm SL}(2,2^p)$ for $p > 3$ with an automorphism of order $p$, so what I said here is not correct)
More examples of a similar nature: if $H$ is a finite non-soluble group and $p$ a prime not dividing $|H|$, let $G$ be $H^p\rtimes \mathbb{Z}/p$, and $x$ an element of order $p$.
In general, if $G$ is finite and $x$ is an element of order $p$ generating a Sylow $p$-subgroup of $G$, then your condition implies that $G$ has a normal subgroup of index $p$, so it looks somewhat like the above examples.
For big infinite simple examples you could take the Tarski monsters. If $G$ is a Tarski monster and $x$ and $y$ are conjugate elements of order $p$, then either $x=y$ or $x$ and $y$ generate $G$.
Looking more deeply at this question, there is quite a long history of related questions, going back long prior to the classification of finite simple groups, for examples in results of B.Fischer which may be viewed as related to later theorems of M. Aschbacher and of G. Glauberman. The classification of the finite simple groups allows much stronger conclusions than were previously available. As @YCor noted in comments, your condition implies that no two distinct elements of $C$ commute.
The famous $Z^{\ast}$-theorem of Glauberman shows in case $C$ is a conjugacy class of involutions, the class $C$ has your property if and only if $G = O_{2^{\prime}}(G)C_{G}(x)$ for each $x \in C.$ More generally, if $C$ is a conjugacy class of elements of prime power order (say $p^{m}$), then a post CFSG theorem of Glauberman, Guralnick, Lynd and Navarro again proves that $C$ has your property if and only if $G = O_{p^{\prime}}(G)C_{G}(x)$ for each $x \in C$. At present, there is no known proof of this fact without CFSG, whereas Glauberman's original result does not require CFSG when $p=2$. The case that $C$ does not consist of elements of prime power order is more difficult. In the case that $G$ acts doubly transitively (by conjugation) on a conjugacy class $C$ of elementz of prime order, then either all elements of $C$ commute with each other, or your condition is satisfied, and in the latter case, some ( pre-CFSG) results of M. Aschbacher and of B. Fischer are relevant. Paul Flavell and I proved the following theorem ( without CFSG, but using character theory) which is mildly relevant to your question, but the hypotheses are too strong: If $x$ is a $\pi$-element of a finite group $G$, then the following are equivalent:
i) $G = O_{\pi^{\prime}}(G)C_{G}(x)$.
ii) Whenever $g \in G$, $\langle x,x^{g} \rangle$ has a normal $\pi$-complement, and furthermore $[\langle x,x^{g} \rangle, x] \cap C_{G}(x) \leq O_{\pi^{\prime}}(C_{G}(x))$ for all $g \in G.$
( In either case, $C = \{x^{G} \}$ satisfies your condition).
I am wondering if this property (which to me sounds quite strong) forces some property on the group $G$, in particular in terms of solvability.
The first solvable example that occurred to me is $D_n$, the dihedral group of order $2n$ where $n$ is odd, with $C$ = the conjugacy class of involutions. The fact that this is an example with the desired property is a special case of the following more general fact.
Fact. Suppose that $G$ is a finite group and $x\in G$ generates a cyclic Sylow $p$-subgroup $P$. The conjugacy class $C = x^G$ has the desired property if and only if the subgroup $P=\langle x\rangle$ has a normal complement.
Reasoning.
Suppose that $P = \langle x\rangle$ is a cyclic
$p$-Sylow subgroup of $G$ and $C = x^G$. If $y\in C\cap \langle x\rangle$
and $y\neq x$, then $y$ is conjugate to $x$ in $G$ but not conjugate
to $x$ in the commutative subgroup $\langle x, y\rangle = \langle x\rangle$.
Thus, if $G$ has the desired property, then
$C\cap \langle x\rangle = \{x\}$. This implies that
$N_G(x) = C_G(x)$ (normalizer of $x$ = centralizer of $x$), which implies that the Sylow subgroup $P=\langle x\rangle$
has a normal complement.
Conversely, assume that the cyclic Sylow subgroup $P=\langle x\rangle$ has a normal complement $N\lhd G$ and $C = x^G$. If $x, y\in C$, then $P=\langle x\rangle$ and $P'=\langle y\rangle$ are conjugate in $\langle x, y\rangle$ by Sylow's Theorem. The element $x$ is therefore conjugate in $\langle x, y\rangle$ to some element of $C\cap \langle y\rangle$. But I claim that $C\cap \langle y\rangle = \{y\}$, forcing $x$ to be conjugate to $y$ in $\langle x, y\rangle$. This claim is justified by the observation that if $z\in C\cap \langle y\rangle$, then the ratio $yz^{-1}$ of of $G$-conjugate elements which lie in $\langle y\rangle$ will belong to $[G,G]|_{\langle y\rangle}\subseteq N|_{\langle y\rangle} = \{1\}$. Here I am using that $N$ is also a complement of $\langle y\rangle$. Since the ration $yz^{-1}$ must be $1$, we get that $z\in C\cap \langle y\rangle$ implies $z=y$. \\\
You can apply this to $D_n$, $n$ odd, by letting $p=2$ and $x$ be an involution.
This property generalizes to the context of self-distributivity, racks, and quandles.
We say that an algebraic structure $(X,*)$ is left-distributive if it satisfies the identity $x*(y*z)=(x*y)*(x*z)$. We say that an algebraic structure $(X,*,*^{-1})$ is a rack if $*$ is left-distributive and $x*(x*^{-1}y)=x*^{-1}(x*y)=y$. If $(X,*,*^{-1})$ is a rack, then $*^{-1}$ is left-distributive as well. If $(X,*)$ is left-distributive or if $(X,*,*^{-1})$ is a rack, then for $a\in X$, we define a mapping $L_a:X\rightarrow X$ by letting $L_a(x)=a*x$. Left-distributivity says that each $L_a$ is an endomorphism. If each $L_a$ is an automorphism, then $(X,*)$ can be endowed with a $*^{-1}$ operation so that $(X,*,*^{-1})$ is a rack. If $(X,*)$ is a left-distributive algebra, then the inner endomorphism monoid $\text{Inn}(X)$ is the monoid generated by $\{L_a:a\in X\}$. If $(X,*,*^{-1})$ is a rack, then $\text{Inn}(X)$ is a group. We say that a rack $(X,*)$ is connected if the group $\text{Inn}(X)$ acts transitively on $X$. We say that a finite rack is hereditarily connected if each subrack of $(X,*)$ is connected. A rack $(X,*)$ is said to be a quandle if it satisfies the identity $x*x=x$.
Proposition: A finite rack $(X,*)$ is hereditarily connected if whenever $x,y\in X$, there are $a_1,\dots,a_n\in\{x,y\}$ and $e_1,\dots,e_n\in\{-1,1\}$ with $(L_{a_1}^{e_1}\circ\dots\circ L_{a_n}^{e_n})(x)=y$.
Proof: The direction $\leftarrow$ is clear. The direction $\rightarrow$ can be established using the identities $L_{a*b}=L_a\circ L_b\circ L_a^{-1}$ and $L_{a*^{-1}b}=L_a^{-1}\circ L_b\circ L_a.$ If $x,y\in X$, then there are $a_1,\dots,a_n$ in the subrack generated by $x,y$ where $(L_{a_1}^{e_1}\circ\dots\circ L_{a_n}^{e_n})(x)=y$. But we can factor each $L_{a_i}$ as a composition of the mappings $L_x,L_y,L_{x}^{-1},L_{y}^{-1}$ using the above identities. Q.E.D.
We observe that if $C$ is a conjugacy class in a group $G$ that generates $G$, then $(C,*,*^{-1})$ is a quandle where $x*y=xyx^{-1},x*^{-1}y=x^{-1}yx$. Furthermore, if $\forall x,y\in C,\exists h\in\langle x,y\rangle:hxh^{-1}=y$, then the quandle $(C,*,*^{-1})$ is hereditarily connected. If $C$ is a subset of a group $G$ which is closed under conjugation by elements in $C$, then let $\Gamma(G,C)=(C,*,*^{-1})$.
Proposition: Suppose that $(X,*,*^{-1})$ is a hereditarily connected rack. Let $G=\text{Inn}(X,*,*^{-1})$. Let $C=\{L_a:a\in X\}$. Then $C$ is a conjugacy class in $G$ where whenever $r,s\in C$, there is some $h\in\langle r,s\rangle$ with $s=hrh^{-1}$.
Proof: Clearly $C$ generates $G$. Suppose now that $x,y\in X$. Then there are $a_1,\dots,a_n\in\{x,y\},e_1,\dots,e_n\in\{-1,1\}$ where $L_{a_1}^{e_1}\circ\dots\circ L_{a_n}^{e_n}(x)=y$. But if we set $h=L_{a_1}^{e_1}\circ\dots\circ L_{a_n}^{e_n}$, then $h\circ L_x\circ h^{-1}=L_y.$ Q.E.D.
If $(X,*,*^{-1})$ is a rack, then let $\Delta(X,*,*^{-1})=(\text{Inn}(X,*,*^{-1}),\{L_x:x\in X\})$.
If $(X,*,*^{-1})$ is a rack, then define a function $j:X\rightarrow\Gamma(\Delta(X))$ by letting $j(x)=L_x$. Then $j$ is a surjective rack homomorphism.
Proposition: If $(X,*,*^{-1})$ is a hereditarily connected quandle, then the mapping $j$ is a quandle isomorphism.
Proof: We only need to show that $j$ is injective. Suppose that $x,y\in X$. If $L_x=L_y$, then $x*x=y*x=x$, but by hereditary connectivity, we can get $x=L_{a_1}^{e_1}\circ\dots\circ L_{a_n}^{e_n}(x)=y$ for some $a_1,\dots,a_n\in\{x,y\},e_1,\dots,e_n\in\{-1,1\}$. Q.E.D.
Suppose now that $G$ is a group and $C\subseteq G$ is closed under conjugation $G$ and generates $G$. Then define a group homomorphism $j:G\rightarrow\Delta(\Gamma(G,C))$ by letting $j(g):C\rightarrow C$ be the permutation where $j(g)(c)=gcg^{-1}$. Then $j$ is a surjective group homomorphism with $\ker(j)=Z(G)$, so $j$ is a group isomorphism if and only if $G$ is centerless.
Proposition: Suppose that $G$ is a group and $C\subseteq G$ is closed under conjugation and generates $G$. Suppose furthermore that $\forall x,y\in C,\exists h\in\langle x,y\rangle,hxh^{-1}=y$. Then $G/Z(G)$ is centerless.
Proof: Suppose that $(G_1,C_1)=\Delta(\Gamma(G,C))$ and $(G_2,C_2)=\Delta(\Gamma(G_1,C_1))$. Then $\Gamma(G,C)$ is a hereditarily connected quandle, so $j:\Gamma(G,C)\rightarrow\Gamma(\Delta(\Gamma(G,C))$ is a quandle isomorphism that induces a group homomorphism $\Delta(j):\Delta(\Gamma(G,C))\rightarrow\Delta(\Gamma(\Delta(\Gamma(G,C))))$. This means that the quotient mapping $\pi:G/Z(G)\rightarrow (G/Z(G))/Z(G/Z(G))$ is a group isomorphism. Therefore, $G/Z(G)$ is centerless. Q.E.D.
We conclude that the mappings $\Gamma,\Delta$ give a duality between hereditarily connected quandles and pairs $(G,C)$ where $G$ is a centerless group and $C$ is a conjugacy class where $\forall x,y\in C\exists h\in\langle x,y\rangle,hxh^{-1}=y$.
