I believe that, if you make a plausible-sounding assumption on the SHA256 hash compression operation, you can show that the only malleability SHA256 has are length extension attacks.
This plausible assumption is that, for a fixed input state, then the mapping between message block and output state acts like a random oracle [1]
With this assumption, suppose we a message $M_0$ which, after SHA-256 padding, is the sequence $N_0$, and a message $M_1 = f(M_0)$, which, after SHA-256 padding, is the sequence $N_1$, and we assume that $M_1$ does not have $M_0$ as a prefix (if it is, then this is a length extension attack).
If $M_0$ does not have $M_1$ as a prefix, then we can show that, after some integral number of SHA-256 blocks, $N_0$ and $N_1$ differ; at this point, the SHA-256 hash compression operation will map their states to random values, and after that point, the latter SHA-256 hash compression operations will continue to map the states to random values, and so the outputs will be effectively random (and hence you cannot compute one from the other).
And, if $M_0$ does have $M_1$ as a prefix (the attempted attack is a "length shortening attack"), then $N_0$ may have $N_1$ as a prefix; if it does, then the intermediate state of $M_0$ processing cannot be determined by the final $M_0$ output value (as it is effectively random), and $N_0$ doesn't have $N_1$ as a prefix, then the previous reasoning applies.
This argument is a bit hand-wavy; I believe that it's essence is valid.
[1]: Note that we cannot make the assumption that the entire hash compression operation acts as a random oracle, because we know how, given a message block $M$ and a delta $\delta$, we can find input and output states $S_0$ and $S_1$ with $S_1 = \text{Compress}( S_0, M )$ and $S_1 = S_0 + \delta$; this shows that the hash compression operation itself is distinguishable from a random oracle.
