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A modification of Guntram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

A modification of Guntram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

Update: in fact, in the above example the cyclic group $\langle a \rangle$ can be replaced with an arbitrary residually finite group $F$. Any residually finite group $F$ is "approximated" by homomorphisms into finite symmetric groups (e.g.,via the natural actions on cosets of finite index subgroups), giving rise to a sequence approximating homomorphisms of $F$ into $Inn(S_n)\cong S_n$ (where $n$ varies). One can define the action of $F$ on the direct product of $\times S_n$ as before and check that the corresponding semidirect product $H=(\times S_n)\rtimes F$ satisfies Drike's property. The base group will consist exactly of the elements with finite conjugacy classes.

Thus any residually finite group can be embedded into a group with the described property. It is easy to see that the property implies that the group is residually finite, so there is no room for improvement.

A modification of Gumtram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

A modification of Guntram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

A modification of Gumtram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not and FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

A modification of Gumtram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.