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 (MSN) contain most of what I have written (and also answer your question).