Do two dimensional representations with the same adjoint representation differ by a character? Let $K$ be a field of characteristic not equal to $2$.  Let $\text{ad} : \text{GL}_2(K) \to \text{GL}_3(K)$ be the adjoint representation, obtained by $\text{GL}_2(K)$ acting on $2 \times 2$ matrices with trace $0$ by conjugation.  Suppose $\rho_1, \rho_2 : G \to \text{GL}_2(K)$ are representations of a group $G$ such that $\text{ad}\rho_1 \cong \text{ad}\rho_2$ over $K$. Is there a character $\eta : G \to K^\times$ such that $\rho_1 \cong \rho_2 \otimes \eta$ over $K$?  That is, if $x \in \text{GL}_3(K)$ such that $\text{ad}\rho_1 = x(\text{ad}\rho_2) x^{-1}$, can one produce $y \in \text{GL}_2(K)$ and $\eta$ as above such that $\rho_1 = y \eta (\otimes \rho_2) y^{-1}$?
I am most interested in the case when $K$ is a finite field.  In that case, I believe one can treat the case when the projective image of $\rho_i$ is isomorphic to $\text{PSL}_2(K)$ or $\text{PGL}_2(K)$ fairly easily by using that those groups have automorphism group $\text{P}\Gamma\text{L}_2(K)$.  But I'd like a proof that does not break into cases based on the subgroups of $\text{PGL}_2(K)$.
 A: $\DeclareMathOperator\ad{ad}$This is true. The way I am viewing it, one has to make two cases, the representation $\ad V$ (and hence $\ad W$) is irreducible, or both are dihedral.  
Part I: Assume that $\ad V$ and $\ad W$ are irreducible ($\rho_1 : G \rightarrow GL(V)$ and $\rho_2:G \rightarrow GL(W)$. Since $V$, $W$ are two dimensional, $V^*=V\otimes \bigwedge^2V^*$, $W^*=W\otimes \bigwedge^2 W^*$. Then $1\oplus \ad V= V\otimes V^*=V\otimes V\otimes \bigwedge^2V*=1\oplus S^2V\otimes \bigwedge^2V^*$. By assumption, this is isomorphic to $W\otimes W^*$, hence we get $S^2V\otimes \bigwedge^2V^*=S^2W\otimes \bigwedge ^2 W^*$, i.e. $S^2V\otimes \bigwedge ^2W= S^2W\otimes \bigwedge ^2V$.
Claim: $V\otimes W=E\oplus F$ is not possible if $E$, $F$ are  two-dimensional representations. For, if such $E$, $F$ exist, then, taking second exterior on both sides, we get 
$$S^2V\otimes \bigwedge^2 W\oplus S^2W\otimes \bigwedge^2V=\bigwedge^2E\oplus \bigwedge^2 F \oplus E\otimes F.$$ Now use the preceding para to conclude that $S^2V\otimes \bigwedge^2W$ contains $\bigwedge^2E$, contradicting irreducibility: $\ad V=S^2V\otimes \bigwedge^2V^*\supset \bigwedge^2W^*\otimes \bigwedge^2 E \otimes \bigwedge^2 V^*$.
Claim: $V\otimes W$ is reducible. For   $V\otimes W\otimes V^*\otimes W^*= (1\oplus \ad V)\otimes (1\oplus \ad W)\supset 1\oplus \ad V\otimes \ad W$ contains a two-dimensional space of invariants: clearly $\ad W=\ad W^*$, and $\ad W^*=\ad V^*$ by assumption. 
The above two claims  imply that $V\otimes W$ contains a one-dimensional invariant subspace $E$. Therefore, $W=V^*\otimes E=V\otimes \bigwedge^2V^*\otimes E$, and thus $W$ is got from $V$ by tensoring with a character. 
Part II : When $\ad V=\ad W$ is not irreducible, $V$, $W$ are dihedral (I have a long, but easy proof) and the equality of $\ad V$ and $\ad W$ implies the equivalence  of $V$ and $W$ by arguments similar to the preceding (it is long, but again easy).  
The first few sections of Murty and Prasad - Tate cycles on a product of two Hilbert modular surfaces (MSN) contain most of what I have written (and also answer your question).
A: Not an answer, but a possible start to a character computation @LSpice, too long for a series of comments. 
Fix $g$ in $G$, and let $a_i$ and $b_i$ be the eigenvalues of $\rho_i(g)$ for $i = 1, 2$. Then the eigenvalues of ${\rm ad} \rho_i(g)$ are $1$, $a_i b_i^{-1}$ and $b_i a_i^{-1}$. By assumption the sets $\{a_1 b_1^{-1}, b_1 a_1^{-1}\}$ and $\{a_2 b_2^{-1}, b_2 a_2^{-1}\}$ are the same. If $a_i b_i^{-1} \neq \pm 1$ then these two sets consist of distinct elements, and up to switching $a_2$ and $b_2$, we can identify that $a_1 b_1^{-1} = a_2 b_2^{-1}$. Therefore, if such a character $\eta$ were to exists, we would have to have $\eta(g) = a_1 a_2^{-1} = b_1 b_2^{-1}$. This is still true if $a_i b_i^{-1} = 1$: now $a_i = b_i$, so we do not have to worry about picking which of the two eigenvalues of $\rho_2(g)$ to call $a_2$, as they are the same. It's only if $a_i b_i^{-1} = -1$ (equivalently, ${\rm tr} \rho_i(g) = 0$) that we do not know which eigenvalue to call $a_2$. But since $b_2 = - a_2$, the quantity $\pm a_2$ is well-defined; in this case $\eta(g) = \pm a_1 a_2^{-1}$ is only determined up to sign. 
So we have now defined a quasi-function $\eta: G\ "\!\to\!"\ K^\times$ with $\eta(g) = a_1 a_2^{-1}$ whenever ${\rm tr} \rho_i (g) \neq 0$ and $\eta(g)$ defined only up to sign otherwise. (I am just using "quasi-function" informally here.) Note that this quasi-function satisfies ${\rm tr} \rho_1 = \eta \cdot {\rm tr} \rho_2$, so that if one could prove that the quasi-function can be refined to a function that is a character $G \to K^\times$, then by Brauer-Nesbitt --- assuming we are free to assume $\rho_i$ are semisimple --- we would be done. 
Moreover, note that $\eta^2: G \to K^\times$ is not only a perfectly well-defined function but also a character: indeed, $\eta^2 = {\rm det} \rho_1 \otimes {\rm \det \rho_2}^{-1}$. 
To be continued, as I must dash, but possibly someone already has an idea for how to take it from here?
Post-dash sequel:  
[At this point, by the way, we have a proof for $K$ of characteristic $2$: there is no sign ambiguity, so that $\eta: G \to K^\times$ is a proper function; and since $\eta^2$ is a character, so is $\eta$. Finally since $\rho_1$ and $\rho_2 \otimes \eta$ have the same trace and determinant, and we are assuming $\rho_i$ semisimple, Brauer-Nesbitt applies to conclude that they are isomorphic representations.] 
Some ideas about how to proceed from here, to see that $\eta$ is multiplicative where defined:
From the definition, it follows that $\eta(1) = 1$.
The identity ${\rm tr} \rho_i(g^{-1}) = \det \rho_i(g)^{-1} {\rm tr} \rho_i(g)$ combined with the trace and determinant equations gives us ${\rm tr} \rho_2(g) \eta(g^{-1}) = {\rm tr} \rho_2(g) \eta(g)^{-1}$, so that $\eta(g^{-1}) = \eta(g)^{-1}$ so long as ${\rm tr} \rho_i(g) \neq 0$. 
Since $\det \rho_i(g) = \frac{1}{2}\big(({\rm tr} \rho_i(g))^2 - {\rm tr} \rho_i(g^2)\big)$ and $\det \rho_1(g) = \det \rho_2(g) \eta(g)$ and  ${\rm tr} \rho_1(g) = {\rm tr} \rho_2(g) \eta(g)$, we can conclude that ${\rm tr} \rho_2(g^2) \eta(g)^2 = {\rm tr} \rho_2(g^2) \eta(g^2)$ for all $g \in G$. Therefore, if ${\rm tr} \rho_i(g^2) \neq 0$, then $\eta(g)^2 = \eta(g^2)$. (I don't know how useful this is.)
The general version of the two identities above is the pseudocharacter identity $${\rm tr} \rho_i(gh) + \det \rho_i(g) {\rm tr} \rho_i(g^{-1} h) = {\rm tr} \rho_i(g) {\rm tr} \rho_i(h),$$ which holds for any $g, h$ in $G$. Whence, for $g, h \in G$, $${\rm tr} \rho_2(gh) \big(\eta(gh) - \eta(g) \eta(h)\big) = \det \rho_2(g) {\rm tr} \rho_2(g^{-1} h) \big(\eta(g) \eta(h) - \eta(g)^2 \eta(g^{-1} h) \big).$$
Can anything useful come out of this? 
