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.



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


