I have a tensor $T_{mn}$ where its indices coorespond to a flat metric $g_{mn}$. I want $T_{mn}$ to be a new metric $\tilde{g}_{mn}$, such that $T_{mn}(g_{rs}) = \tilde{g}_{mn}$.
A theorem says that all 2D metrics are conformally flat. So does their exists a conformal factor $\phi$, such that \begin{align} T_{mn}(g_{rs}) = \phi g_{mn} && (1) \end{align} is possible? If so, does this imply that $T_{n}{}^{n}{}=\phi$?
The confusing element in this problem is $T_{mn}$'s indices correspond to the indices of the flat metric $g_{mn}$ from which it was created. The above may be incorrect, and a new metric flat metric $h_{mn}$ must be created to allow \begin{align} T_{mn}(g_{rs}) = \phi h_{mn} && (2) \end{align} where $h_{mn}\neq g_{mn}$.
Are all 2D tensors in a specified flat metric equal to that same metric conformally scaled?
Okay, I think I understand enough of your question to give an answer.
And that is because two flat metrics on the same 2D manifold do not have to be conformal. To wit, let $\mathbb{T}^2$ denote the torus. The metrics with line elements
$$ \mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 $$
and
$$ \mathrm{d}s^2 = \mathrm{d}x^2 + 2 \mathrm{d} y^2 $$
are both flat, but are not conformal to each other.