Permutation of keys that guarantees different hashes Suppose I have two values to be hashed, $x$ and $y$. Is there a hash function $H$ with the following properties?

$H(x) = H(y), \; H(m(x)) \neq H(m(y))$

Where $m$ is a permutation function that may depend on $H$.

Edit: The corrected question is, if $H(x) = H(y)$, them is it possible to make any permutation $m$ so that $H(m(x)) \neq H(m(y))$?
I assume you're asking whether there exists a hash function $H: \{0,1\}^* \to \{0,1\}^\ell$, and a permutation $m: \{0,1\}^* \to \{0,1\}^*$, such that $$H(x) = H(y) \implies H(m(x)) \ne H(m(y))$$ for all $x \ne y$.

The answer, alas, is no &mdash; there is no such $H$ and $m$.

If there was, we could construct a collision-free hash function $H': \{0,1\}^* \to \{0,1\}^{2\ell}$ as $$H'(x) = H(x) \| H(m(x)).$$  But since the set $\{0,1\}^*$ is infinite, while $\{0,1\}^{2\ell}$ is finite, this is clearly impossible.

More generally, the same argument shows that no such $H$ and $m$ can exist if the input domain of $H$ has more than $n^2$ elements, where $n = 2^\ell$ is the number of possible outputs of $H$.