In the paper ``Morse theory on Hilbert manifolds'' (1963), on page 326, Richard Palais makes a remark that if $\phi\colon V \to W$ is an isometry (of submanifolds of $\mathbb{R}^n$), then this does not imply that the natural lift $\bar{\phi}\colon H_1([0,1];V) \to H_1([0,1];W)$ is an isometry of Hilbert manifolds. There is no explanation of why this is not generally true, though. This remark is puzzling me, since I don't see how this would fail to be true.

A little bit of background details. Palais constructs the Hilbert manifold $H_1([0,1];V)$ non-intrinsically (as compared to the construction in Klingenberg's ``Riemannian geometry'') through the embedding of $V$ into some $\mathbb{R}^n$ and then choosing a metric on $\mathbb{R}^n$ such that $V$ is a totally geodesic submanifold.

So my question is: what is wrong with my argument below that $\bar{\phi}$ is indeed an isometry?

Let $V,W$ be two embedded submanifolds in Euclidean space, so $V,W$ become Riemannian manifolds with metric given by the restriction of the Euclidean metric to their respective tangent bundles. Let $\phi\colon V \to W$ be an isometry.

Now let $\sigma \in H_1([0,1];V)$ and $\lambda,\mu \in H_1([0,1];V)_\sigma$. The metric $g$ on $H_1([0,1];V)$ is defined by $$ g_\sigma(\lambda,\mu) = \int_0^1 \langle \lambda(t), \mu(t) \rangle \,dt $$ with $\langle \,\cdot\,, \,\cdot\, \rangle$ the Euclidean inner product. Thus, with $\bar{\phi}(\sigma) = \phi \circ \sigma$ we get (using $d\bar{\phi}$ as given in Theorem~7 of the paper) $$ \bar{\phi}_*(g)_{\bar{\phi}(\sigma)}(d\bar{\phi}_\sigma(\lambda), d\bar{\phi}_\sigma(\mu)) = \int_0^1 \langle d\phi_{\sigma(t)}(\lambda(t)), d\phi_{\sigma(t)}(\mu(t)) \rangle \,dt. $$ Since $\phi$ was an isometry, this reduces to the metric on $H_1([0,1];W)$.

Note that I interpreted $H_1([0,1];V)$ and $H_1([0,1];W)$ as Riemannian manifolds, where I assumed that the metric is defined by the inner product on the canonical Hilbert spaces on which the tangent spaces are modeled.