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 $\text{PGL}\_2(K)$.



Do two dimensional representations with the same adjoint representation differ by a character?
$\DeclareMathOperator\ad{ad}$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](https://www.sciencedirect.com/science/article/pii/S0022314X99924460) ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=1735646)) contain most of what I have written (and also answer your question).


