The HHL algorithm generally can be thought of as diagonalizing our matrix $A$ with the quantum phase estimation algorithm, and applying a specific function $f(\lambda)=\lambda^{-1}$ to the eigenvalues so obtained by rotating an ancilla qubit and post-selecting on the ancilla qubit being $|1\rangle$. The post-selection probability is given by the condition number $\kappa$, or the difference between the largest and smallest eigenvalues of $A$. We then kick these reciprocal eigenvalues back to our wavefunction by uncomputing the quantum phase estimation.
But after we diagonalize $A$, I don't think anything precludes us from applying any (Lipschitz continuous?) one-to-one function $f$ to our eigenvalues. Indeed I can imagine that we could apply $f(\lambda)=\lambda^m$ for some positive integer $m$, and then kick these back to our state.
This feels as if we can prepare a properly normalized version of $A^m|\psi\rangle$ using a variant of the HHL algorithm (which would normally evaluate $A^{-1}|\psi\rangle$) if we were successful in post-selecting our ancilla. But, if $A$ were the adjacency matrix of a connected undirected graph, and if $m$ were large enough, then by the Perron-Frobenius theorem I think we can also say that $A^m|\psi\rangle$ is indeed the stationary distribution, or the ground state, of $A$. Clearly we can't easily prepare the ground state for any arbitrary hermitian matrix, as this is QMA-hard.
So, under what conditions could we prepare $A^m|\psi\rangle$ using a version of HHL where the reciprocal $\lambda^{-1}$ rotation is replaced with positive exponentiation $\lambda^m$? Would then the post-selection probability given by the spectral gap (the largest vs. second-largest eigenvalue) in lieu of the condition number (the largest vs. smallest eigenvalue)?