Isometric deformation of a helicoid with $n$ turns

Question

Given a helicoid, $\mathcal{H}$, of pitch $P$ with $n$ turns, where $$n = \dfrac{L}{P},$$ can I not isometrically deform $\mathcal{H}$ into a helicoid with just one turn by varying the length $L$ of the axis so as to keep $P$ fixed. The fundamental forms are all dependent on $P$ only, so the resulting surfaces will be congruent. But by varying $L$, am I not stretching and thus disturbing the arc length element? This is probably a nonsensical observation or I am missing something simple. I wanted to verify however.