General linear group of degree two is ambivalent implies multiplicative group of ring is trivial

Statement
Suppose $$R$$ is a commutative unital ring. Then, if the general linear group of degree two $$GL(2,R)$$ is an ambivalent group (i.e., every element is conjugate to its inverse), the multiplicative group of $$R$$ is a trivial group. In the case that $$R$$ is a field, this happens if and only if it is field:F2, and in that case, the converse also holds.

Case of a ring
Suppose $$R$$ is a commutative unital ring with a nontrivial multiplicative group. In other words, there exists an invertible element $$u \ne 1$$ in $$R$$.

For a matrix with characteristic polynomial $$x^2 + lx + m$$, the inverse has characteristic polynomial $$x^2 + (l/m)x + (1/m)$$ (basically, we reverse the order of coefficients and then normalize). In order for the matrices to not be conjugate, we need these characteristic polynomials to differ. This can work as long as $$m \ne 0,1$$ and $$l \ne 0$$. The choice $$m = l = u$$ works, and we can construct matrices:

$$A = \begin{pmatrix} 0 & -u \\ 1 & u \\\end{pmatrix}, A^{-1} = \begin{pmatrix} 1 & 1 \\ -1/u & 0 \\\end{pmatrix}$$

These are clearly inverses of each other. However, they do not have the same trace, so they are not conjugate in $$GL(2,R)$$.

Case of a field
Since the multiplicative group of a field comprises all the nonzero elements, the multiplicative group is trivial if and only if the field is field:F2. In this case, the group is symmetric group:S3, which is indeed an ambivalent group.

Rings that are not fields but have trivial multiplicative group
Note that any such ring must have characteristic two, otherwise $$-1$$ is a non-identity element in the multiplicative group. Examples of rings with trivial multiplicative group are:


 * Direct products (finite or infinite) of copies of field:F2.
 * The polynomial ring $$\mathbb{F}_2[t]$$.

The result here does not allow us to conclude anything about whether the groups $$GL(2,R)$$ over such rings are ambivalent. The subtlety arises because, although every element and its inverse have the same characteristic polynomial, having the same characteristic polynomial does not imply being conjugate when we are working over an arbitrary commutative unital ring, although the converse implication (namely, conjugate elements have the same characteristic polynomial) does hold.