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This is only the partial answer: a function $f$ such that $f(⋅−k)$ form an orthonormal basis with the same span as $(g(⋅−λ_k))$ exists only in very special cases. Assume it exists, then on the Fourier transform side span $(g-\lambda_k)$ is the space of functions of the form $\phi\cdot \hat f$ where $\phi$ is $2\pi$-periodic, so in particular we have $e^{i\lambda_k t}\hat g(t)=\phi_k(t)\hat f(t)$. Let $S$ be the support of $\hat g$. Clearly $\hat f$ is non zero on $S$, so on $S$ we have $e^{i\lambda_k t}=\phi_k(t)[\hat f(t)/\hat g(t)]$. Writing this for integers $k\neq s$ and dividing we see that $e^{i(\lambda_k-\lambda_s)t}$ is $2\pi$ periodic on $S$. If the measure of $S$ is $>2\pi$ there are two sets of positive measure $A$ and $B$ such that $A=2\pi \mu +B$ for some $\mu\in Z$. Thus for $t\in B$ we have $e^{i(\lambda_k-\lambda_s)(t+2\pi \mu)}=e^{i(\lambda_k-\lambda_s)t}$ which gives $e^{i(\lambda_k-\lambda_s)2\pi \mu}=1$ so $\lambda_k-\lambda_s$ is an integer for all $k,s\in Z$ which implies $\lambda_k=k+\delta$. If measure of $S$ is $\leq2\pi$ then such $f$ can exist only if $\sum_{s\in Z}I(t+2\pi s t)=1$ for $I$ an indicator function of $S$.