Certainly you can recover $f$ for a circle, because the group is abelian. In general, $f$ can be written as a linear combination of matrix coefficients, and for a general compact Lie group you take a trace, so you will loose information about $f$.

For example, there are nonzero functions with vanishing Fourier transform in your case, e.g. take two orthogonal vectors in an irreducible subspace $V_\rho$ and consider the function given by the matrix coefficient $f:k \mapsto \langle v_1 , \rho(k) v_2 \rangle$. We have $\hat{f}( \rho')=0$ for all irreducible representations $\rho'$. Sketch of proof: Use that $tr\rho'(f) = tr\rho'(f^G)$ where $$f^G (x) = \int\limits_G f(g^{-1}xg) d g$$ and use the orthogonality relations to show that $f^G=0$.

But you can always check the support at the identity with the Plancherel formula!

Certainly you can recover $f$ for a circle, because the group is abelian. In general, $f$ can be written as a linear combination of matrix coefficients, and for the Fourier transform of a general compact Lie group you take a trace, so you will loose information about $f$.

For example, there are nonzero functions with vanishing Fourier transform in the non-commutative situation, e.g. take two orthogonal vectors in an irreducible subspace $V_\rho$ and consider the function given by the matrix coefficient $f:k \mapsto \langle v_1 , \rho(k) v_2 \rangle$. We have $\hat{f}( \rho')=0$ for all irreducible representations $\rho'$. Sketch of proof: Use that $tr\rho'(f) = tr\rho'(f^G)$ where $$f^G (x) = \int\limits_G f(g^{-1}xg) d g$$ and use the orthogonality relations to show that $f^G=0$.

But you can always check the support at the identity with the Plancherel formula $$ f(1) = \sum\limits_{\rho} C_\rho \hat{f}(\rho).$$ The fudge factor $C_\rho$ depend only on the dimension. Unfortuantely, I don't recall how.