Well, if there are regions on the real line where the processes never get (which can be the case, e. g. if $f\equiv 1$ and $\sigma$ has zeroes with fast enough decay), then, clearly, you cannot say anything about $\sigma$'s there.

Othervise the quadratic variation is the invariant that ensures $\sigma=\pm\tilde{\sigma}$. Suppose that $|\sigma(x)| <a<b<|\tilde{\sigma}(x)|$ and let $\tau_x=\min\{t:X_t=x\}$. If $\tau_x<\infty$ with positive probability, then for a small $\epsilon>0$ also $|\sigma(X_t)| <a<b<|\tilde{\sigma}(X_t)|$ for $\tau_x<t<\tau_x+\epsilon$ with positive probability. But on this event $$\langle X_t\rangle_{[\tau_x,\tau_x+\epsilon]}=\int_{\tau_x}^{\tau_x+\epsilon}\sigma^2(X_t)dt<a^2\epsilon$$ and similarly $\langle Y_t\rangle_{[\tau_x,\tau_x+\epsilon]}>b^2\epsilon$, which is a contradiction.

Well, if there are regions on the real line where the processes never get (which can be the case, e.g. if $f\equiv 1$ and $\sigma$ has zeroes with fast enough decay), then, clearly, you cannot say anything about the $\sigma$'s there.

Otherwise the quadratic variation is the invariant that ensures $\sigma=\pm\tilde{\sigma}$. Suppose that $|\sigma(x)| <a<b<|\tilde{\sigma}(x)|$ and let $\tau_x=\min\{t:X_t=x\}$. If $\tau_x<\infty$ with positive probability, then for a small $\epsilon>0$ also $|\sigma(X_t)| <a<b<|\tilde{\sigma}(X_t)|$ for $\tau_x<t<\tau_x+\epsilon$ with positive probability. But on this event $$\langle X_t\rangle_{[\tau_x,\tau_x+\epsilon]}=\int_{\tau_x}^{\tau_x+\epsilon}\sigma^2(X_t)dt<a^2\epsilon$$ and similarly $\langle X_t\rangle_{[\tau_x,\tau_x+\epsilon]}>b^2\epsilon$, which is a contradiction.