# Intuition for why a group can fail to have an automorphism sending a particular element to its inverse.

Below is an arguably geometric example, it may not be what you're looking for.

For a vector space $$V$$ over a field $$F$$, let $$A(V)$$ be the group of affine maps on $$V$$ which are compositions of translations and scaling maps, i.e. the affine maps of the form $$x \mapsto ax + v$$, for $$a \in F \setminus \{0\}$$ and $$v \in V$$. Let $$T(V)$$ be the subgroup of $$A(V)$$ consisting of translations, i.e. the maps $$x \mapsto x + v$$. Note that $$T(V) \cong V$$ as groups. We include the translations because they encode additional structure of the action of scaling maps on $$V$$; if we only considered the group of scaling maps, this would just be $$F \setminus \{0\}$$ under multiplication, which is abelian, and thus has automorphism $$g \mapsto g^{-1}$$.

We will now prove that when $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, every automorphism $$\phi$$ of $$A(V)$$ preserves scaling factor, meaning if $$f \in A(V)$$ is of the form $$x \mapsto ax + v$$, then $$\phi(f)$$ is of the form $$x \mapsto ax + v'$$. In particular, this will mean that every element of $$A(V)$$ with scaling factor $$a \neq \pm 1$$ is not sent to its inverse by any automorphism of $$A(V)$$.

Lemma 1: Let $$m, n$$ be integers which are not zero in $$F$$. Then for $$f \in A(V)$$, $$f$$ has the property that $$fg^n = g^m f$$ for all $$g \in T(V)$$ if and only if $$f$$ has scaling factor $$m/n$$, meaning $$f$$ is of the form $$x \mapsto (m/n)x + v$$.

Proof: If $$f$$ is of the given form, then for $$g(x) = x + u$$, clearly $$f(g^n(x)) = (m/n)(x + nu) + v = (m/n)x + v + mu = g^m(f(x))$$ so $$fg^n = g^m f$$. In the other direction, if $$f$$ satisfies $$fg^n = g^m f$$ for all $$g \in T(V)$$, then writing $$f(x) = ax + v$$, and taking $$g(x) = x + u$$ for some $$u \neq 0$$, we have $$a(x + nu) + v = ax + v + mu$$, giving $$anu = mu$$, hence $$a = m/n$$ as desired.

Lemma 2: If $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, then every automorphism $$\phi$$ of $$A(V)$$ preserves $$T(V)$$, meaning $$\phi(T(V)) = T(V)$$.

Proof: In the case $$F = \mathbb{Q}$$, we can identify the subgroup $$T(V)$$ of $$A(V)$$ as the set of "divisible" elements, namely those elements $$g \in A(V)$$ for which, for any positive integer $$n$$, there is an element $$h \in A(V)$$ with $$h^n = g$$. Elements $$g \in A(V)$$ of the form $$x \mapsto ax + v$$ for $$a \neq 1$$ do not have this property, since there can be an $$h$$ with $$h^n = g$$ only if $$a$$ is an $$n$$-th power in $$\mathbb{Q}$$, and the only nonzero element of $$\mathbb{Q}$$ which is an $$n$$-th power for each $$n$$ is $$a = 1$$. It is clear that any automorphism maps divisible elements to divisible elements, and non-divisible elements to non-divisible elements, so any $$\phi$$ has $$\phi(T(V)) = T(V)$$.

In the case $$F = \mathbb{F}_p$$, we can identify $$T(V)$$ as the set of elements of order $$p$$ (together with the identity). For an element $$g$$ of the form $$x \mapsto ax + v$$ for $$a \neq 1$$, $$g$$ is conjugate to the map $$x \mapsto ax$$, and thus $$g$$ has order dividing $$p-1$$, since this latter map has order dividing $$p-1$$. Automorphisms preserve order, so again any $$\phi$$ has $$\phi(T(V)) = T(V)$$. [end proof]

Now, let $$\phi$$ be an automorphism of $$A(V)$$, and let $$f \in A(V)$$, so $$f(x) = ax + v$$ for some $$a, v$$. Since $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, $$a$$ is of the form $$m/n$$ for some integers $$m, n$$ which are not zero in $$F$$, and thus by Lemma 1, $$fg^n = g^m f$$ for any $$g \in T(V)$$. By Lemma 2, $$\phi$$ preserves $$T(V)$$, so $$\phi(f)$$ also has the property that $$\phi(f) g^n = g^m \phi(f)$$ for all $$g \in T(V)$$. Then, again by Lemma 1, this means that $$\phi(f)$$ has scaling factor $$m/n = a$$, so $$\phi$$ preserves the scaling factor of $$f$$.