The reason is two-fold: 1) A unitary transformation does preserve the norm $\left\|\psi \right\| = \langle \psi | \psi \rangle$, but not only the norm. 2) A quantum measurement must produce a state that is not affected by a repeat identical measurement.
In general, a unitary transformation $U$, $U^\dagger U = U U^\dagger = I$, preserves overlaps:
\langle U\phi | U\psi\rangle = \langle \phi |U^\dagger U |\psi \rangle = \langle \phi | \psi \rangle
Say a normalized state reads $a\psi + b\phi$ before collapse, for orthogonal and normalized $\psi$ and $\phi$, $\langle \phi | \psi \rangle = 0$, $\langle \psi|\psi \rangle = \langle \phi|\phi \rangle = 1$, and $|a|^2 + |b|^2 = 1$ such that $\left\|a\psi + b\phi \right\| = \langle a\psi + b\phi| a\psi + b\phi \rangle = 1$. Let $a\psi + b\phi$ collapse into $\phi$ upon measurement. By definition, a 2nd identical measurement must leave $\phi$ unchanged. If the collapse were a unitary evolution such that $U(a\psi + b\phi) = \phi$, then the same measurement on $\phi$ would have to result in $U\phi = \phi$. The unitary $U$ would indeed preserve the norm, since $\left\|a\psi + b\phi \right\| = \left\|U(a\psi + b\phi) \right\| = \left\|\phi \right\| = 1$. But $U$ should also preserve the overlap $\langle \phi | a\psi + b \phi \rangle = b$, whereas instead
\langle U\phi | U(a\psi + b \phi) \rangle = \langle \phi | \phi \rangle = 1 > b
Since we already took care of normalizations, there is no way to remove the above disagreement by a rescaling of $\phi$. So collapse cannot be unitary.
A faster way to arrive at the same conclusion is to consider collapse from a mixed initial state $\rho$, $\rho \neq \rho^2$. The result of the collapse would still be a pure state, so in this case $U$ would have to take a mixed state into a pure state. But unitary transformations always take pure states into pure states, so again this cannot work.