That's simply wrong; it's perfectly clear what he's asking. He "defines" two Hilbert spaces, $H_1=L^2$ with the standard norm and $H_2=L^2$ with a different norm, and he wants to construct an isometry between the two.
It's clear what he's asking, the problem is that the second norm is simply not a norm on $L^2$, so the question is based on an invalid premise. So an answer pointing out the problem and then correcting the question and answering the corrected version seems appropriate.
There's an answer that was posted before the question was closed. The first version of that answer was wrong, purporting to answer the question as stated. I pointed this out in a comment, and the guy's willing to fix it. But his first revised answer is also wrong, for slightly more subtle reasons.
The point is that the second norm makes a different space into a Hilbert space. The correct definition of that different space involves some subtleties. Trying to get the existing answer fixed by the answerer editing in response to my comments is like pulling teeth - if I could simply post an answer it would work much better. It would be an answer that people who care about such things but don't understand them very well would find interesting and educational:
Sketch The point is that, speaking very informally, the second space should consist of the $f$ such that $f,f'\in L^2$. Here's where it gets interesting and subtle: What people typically mean when they say $f,f'\in L^2$ is not literally $f,f'\in L^2$. (The problem with the current version of the existing answer is that it takes $f,f'\in L^2$ literally, which simply doesn't work.)
In fact "$f,f'\in L^2$" is typically (and here should be) shorthand for any of a long list of equivalent conditions. Seems to me a proper answer really should include at least the statement of a long TFAE theorem - we're never going to get there via the present comment-edit-comment-edit cycle.
So I should vote to reopen? Did that. I gather there exist $n$ such that $n-1$ more reopen votes would do it (hint)...