(This question came up in a conversation with my professor last week.)


Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$.
  

Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ?
  

What if $G$ is finite?



element algebraically distinguishable from its inverse
The Mathieu group $M\_{11}$ does not have this property. A quote from Example 2.16 in [this paper](http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3895v1.pdf): "Hence there is no automorphism of $M\_{11}$ that maps $x$ to $x^{−1}$."


Background how I found this quote as I am no group theorist: I used Google on "groups with no outer automorphism" which led me to this [Wikipedia article](http://en.wikipedia.org/wiki/Outer_automorphism_group), and from there I jumped to this other [Wikipedia article](http://en.wikipedia.org/wiki/Mathieu_group). So I learned that $M\_{11}$ has no outer automorphism. Then I used Google again on "elements conjugate to their inverse in the mathieu group" which led me to the above mentioned paper. 


**EDIT:** Following Geoff Robinson's comment let me show that any element $x\in M\_{11}$ of order 11 has this property, using only basic group theory and the above [Wikipedia article](http://en.wikipedia.org/wiki/Mathieu_group). The article tells us that $M\_{11}$ has 7920 elements of which 1440 have order 11. So $M\_{11}$ has 1440/10=144 Sylow 11-subgroups, each cyclic of order 11. These subgroups are conjugates to each other by one of the Sylow theorems, so each of them has a normalizer subgroup of order 7920/144=55. In particular, if $x$ and $x^{-1}$ were conjugate to each other, then they were so by an element of odd order. This, however, is impossible as any element of odd order acts trivially on a 2-element set.


