Injecting a group into itself

Let $G$ be a group with elements $g_1, g_2\in G$ and injective endomorphisms $\phi_1, \phi_2$ s.t. $\phi_1 (g_1)=g_2$ and $\phi_2(g_2)=g_1$. Does this imply there is an automorphism $\psi$ with $\psi(g_1)=g_2$. I expect the answer to be no but I'm unsuccessful producing a counter example.

Solution 1:

Let $S$ be the group of finitary permutations of $\mathbb{N}$ (i.e., permutations that fix all but finitely many elements of $\mathbb{N}$), and $A$ the subgroup of even permutations. Let $\sigma\in A$ be the $3$-cycle $(1,2,3)$.

There is an injective homomorphism $\theta: S\to A$ with $\theta(\sigma)=\sigma$. [Take a permutation $\tau\in S$ to a permutation of $\mathbb{N}\setminus\{4,5\}$, multiplied by the $2$-cycle $(4,5)$ if $\tau$ is odd.]

Let $G=S\times A$, and $\varphi:G\to G$ the injective homomorphism given by $\varphi(\tau_1,\tau_2)=\left(\tau_2,\theta(\tau_1)\right)$. Then $\varphi(\sigma,1)=(1,\sigma)$ and $\varphi(1,\sigma)=(\sigma,1)$.

However, there is no automorphism of $G$ taking $(\sigma,1)$ to $(1,\sigma)$, since the quotient of $G$ by the normal subgroup generated by $(\sigma,1)$ is isomorphic to $C_2\times A$, whereas the quotient by the normal subgroup generated by $(1,\sigma)$ is isomorphic to $S$, and $C_2\times A\not\cong S$ (only one has non-trivial centre).