This has been incredibly helpful. Thank you!

I've almost understood everything. I'm still a bit hung up on why $\text{ad} V$ being reducible implies $V$ is dihedral. I agree we have $S^2 V$ has a one dimensional invariant subspace. Can you give a few more details about how to get this quadratic form $Q$ and why $\text{GO}(2)$ reps are dihedral?

@Venkatarama: Of course, thank you.

@Venkataramana: in the proof that $V \otimes W$ is reducible, it seems that the key point is that $(V \otimes W) \otimes (V \otimes W)^*$ contains two copies of the trivial representation (under the assumption that $\text{ad}V \cong \text{ad}W$). How does one see that this cannot happen if $V \otimes W$ is irreducible?

Thanks for your answer! This looks like exactly the kind of argument I was looking for. I'm a bit overwhelmed at the moment, so it may take me a week or so for me to think through it carefully. I'm going to leave the post open until I do that in case I have follow-up questions, but hopefully by the end of next week I'll be comfortable accepting this answer. Thanks for your understanding.

I think the answer is easy when both $\rho_i$ are reducible. If $\rho = \chi_1 \oplus \chi_2$ then $\text{ad}\rho = \chi_1\chi_2^{-1} \oplus 1 \oplus \chi_1^{-1}\chi_2$. Using this, I think it's easy to see that $\rho_1$ and $\rho_2$ are twists of each other under the hypothesis that $\text{ad}\rho_1 = \text{ad}\rho_2$.

Certainly the question may be asked in characteristic $2$ as well. I left that out since it is not so relevant to me at the moment, and since I believe I have a proof that is specific to characteristic $2$.