Character-conjugate not implies locally conjugate

Statement
There are actually two statements here:


 * Character-conjugate representations need not be locally conjugate
 * Character-conjugate conjugacy classes need not be locally conjugate

Partial truth
Note that for a finite group and a field whose characteristic does not divide the order of the group, character-conjugate actually is the same as locally conjugate. So the counterexamples must be modular representations.

Proof of the statement for conjugacy classes
Consider the cyclic group $$\mathbb{Z}/p\mathbb{Z}$$ which we denote as $$C_p$$. Let the field be the prime field on $$p$$ elements, which we denote by $$F_p$$.

We claim that any two elements in $$C_p$$ are character-conjugate over $$F_p$$. Let $$g \in C_p$$ and $$\rho$$ be a finite-dimensional linear representation of $$C_p$$ over $$F_p$$. Then, since $$g^p = e$$, $$\rho(g)$$ satisfies the equation:

$$\rho(g)^p - 1 = 0$$

The minimal polynomial of $$\rho(g)$$ thus divides $$x^p - 1 = (x-1)^p$$ and hence all the eigenvalues are $$1$$. Thus the matrix can be conjugated to an upper triangular unipotent matrix. So its trace is the degree of the representation, and is independent of $$g$$.

Clearly, however, it is not true that any two elements of $$C_p$$ are locally conjugate over $$F_p$$. In fact, for any nontrivial representation, the conjugacy class of the identity is different from that of the others. For instance, consider the 2-dimensional representation:

$$a \mapsto \left(\begin{array}{rr}{1}{a}{0}{1}\end{array}\right)$$

Proof of the statement for representations
We need to prove that there can be representations that are character-conjugate, but not locally conjugate. In other words, there can be distinct representations that define the same character, but are not locally conjugate.

The example is similar in spirit to the above. Take the following two two-dimensional representations of $$C_p$$ over $$F_p$$:


 * The trivial two-dimensional representation:

$$a \mapsto \left(\begin{array}{rr}{1}{0}{0}{1}\end{array}\right)$$


 * The following two-dimensional representation:

$$a \mapsto \left(\begin{array}{rr}{1}{a}{0}{1}\end{array}\right)$$

Clearly, these two representations have the same character (the constant function sending everything to 2) but are not locally conjugate.