When do two maps between groups give the same map between representation rings?

Question

If I have a homomorphism $f: G\to H$ of groups, I get a homomorphism $R(f)\colon R(H)\to R(G)$ of representation rings in the opposite direction, by composition. Given two homomorphisms $f_1,f_2\colon G\to H$, it is sometimes the case that $R(f_1)=R(f_2)$. For example, this happens whenever $f_1$ and $f_2$ are conjugate, i.e. there exists $h\in H$ such that, for all $x\in G$, $f_1(x)=h f_2(x) h^{-1}$.

Question: Is it true that if $R(f_1)=R(f_2)$ then $f_1$ and $f_2$ are conjugate? (Or what's a nice counterexample?)

Closely related: Suppose that $f_1(x)$ is conjugate to $f_2(x)$ for each $x\in G$ (but the way they are conjugate a priori depends on $x$). Is $f_1$ in fact conjugate to $f_2$?

The questions make sense pretty generally, but I am mostly interested in, say, finite groups and complex representation rings, if it makes any difference.

Answer 1

Your two questions are not just closely related but identical. If $R(f_1)=R(f_2)$ then $\chi(f_1(x))=\chi(f_2(x))$ for all characters $\chi$, and standard representation theory allows you to deduce that $f_1(x)$ is conjugate to $f_2(x)$.

For a basic example where pointwise conjugacy is different from conjugacy, let $G$ be elementary abelian of order $4$ with generators $a$ and $b$, and let $H$ be the symmetric group on six letters. Define \begin{align*} f_1(a) &= (1\;2)(5\;6) \\\\ f_1(b) &= (3\;4)(5\;6) \\\\ f_1(ab) &= (1\;2)(3\;4) \\\\ f_2(a) &= (1\;2)(3\;4) \\\\ f_2(b) &= (1\;3)(2\;4) \\\\ f_2(ab) &= (1\;4)(2\;3). \end{align*} Note that $5$ and $6$ are fixed by the image of $f_2$, but no point is fixed by the image of $f_1$, so $f_1$ and $f_2$ are not conjugate. However, for each $x$ we see that $f_1(x)$ and $f_2(x)$ have the same cycle type and so are conjugate.