Free cyclic subgroups in a non-abelian group

Question

Is there any non-abelian group $G$ such that for each $a\in G$ and any automorphism $g:\left<a\right>\to \left<a\right>$ the function $$f:G\to G$$ $$f(x) = \begin{cases} g(x) & \text{ if } x\in \left<a\right> \\ x & \text{ if } x\notin \left<a\right> \end{cases}$$ is an automorphism?

Answer 1

Pick $a\in G$ for which $\langle a\rangle$ has a nontrivial automorphism $g$.

Let $b\not\in\langle a\rangle$. What's $f(ab)$? What's $f(a)f(b)$? Can these be equal?

The issue here is that complements of a subset closed under an operation need not themselves be closed, for instance irrational + irrational can be rational, so defining a homomorphism piecewise in this manner is almost surely doomed to fail.