I would say: the free group on one element. I guess you can translate this into a series of first-order axioms. Notice that multiplication comes for free as composition between automorphisms of the group with itself.

Addendum: Prompted by the comment below, I am not thinking about the usual description of the free group through a chain of $1$'s and $-1$'s but on the universal property.

Let me give some specifics. A group is a tuple $(G,m,e,i)$ with $G$ a set, $m \colon G \times G \to G$ a map $e \in G$ and $i \colon G \to G$ satisfying certain commutativities that amount to the defining properties of group (associativity, $e$ is the neutral element and $i(g)$ is the inverse of the element $g \in G$). A free group in one element is such a tuple $(F, \dot , 1, op)$ satisfying that for any choice of a $g \in G$ from a group $(G,m,e,i)$ there is one and only one homomorphism $(F, \dot , 1, op) \to (G,m,e,i)$ taking $1$ to $g$. I propose to translate this description into a series of first order formulas, that was my suggestion.

I would say: the free group on one element. I guess you can translate this into a series of first-order axioms. Notice that multiplication comes for free as composition between automorphisms of the group with itself.

Addendum: Prompted by the comment below, I am not thinking about the usual description of the free group through a chain of $1$'s and $-1$'s but on the universal property.

Let me give some specifics. A group is a tuple $(G,m,e,i)$ with $G$ a set, $m \colon G \times G \to G$ a map $e \in G$ and $i \colon G \to G$ satisfying certain commutativities that amount to the defining properties of group (associativity, $e$ is the neutral element and $i(g)$ is the inverse of the element $g \in G$). A free group in one element is such a tuple $(F, \dot , 1, op)$ satisfying that for any choice of a $g \in G$ from a group $(G,m,e,i)$ there is one and only one homomorphism $(F, \dot , 1, op) \to (G,m,e,i)$ taking $1$ to $g$. I propose to translate this description into a series of first order formulas, that was my suggestion.

Addendum 2: I have just realized that this way the description is second-order.

I would say: the free group on one element. I guess you can translate this into a series of first-order axioms. Notice that multiplication comes for free as composition between automorphisms of the group with itself.

I would say: the free group on one element. I guess you can translate this into a series of first-order axioms. Notice that multiplication comes for free as composition between automorphisms of the group with itself.

Addendum: Prompted by the comment below, I am not thinking about the usual description of the free group through a chain of $1$'s and $-1$'s but on the universal property.

Let me give some specifics. A group is a tuple $(G,m,e,i)$ with $G$ a set, $m \colon G \times G \to G$ a map $e \in G$ and $i \colon G \to G$ satisfying certain commutativities that amount to the defining properties of group (associativity, $e$ is the neutral element and $i(g)$ is the inverse of the element $g \in G$). A free group in one element is such a tuple $(F, \dot , 1, op)$ satisfying that for any choice of a $g \in G$ from a group $(G,m,e,i)$ there is one and only one homomorphism $(F, \dot , 1, op) \to (G,m,e,i)$ taking $1$ to $g$. I propose to translate this description into a series of first order formulas, that was my suggestion.