Such a map will preserve orthogonality, and any such map must be a scalar multiple of an isometry. This is true in great generality, e.g. the map $T$ doesn't have to be linear, and $U$ and $V$ don't have to be finite-dimensional; see Theorem 1 in
Chmieliński, Linear mappings approximately preserving orthogonality. J. Math. Anal. Appl. 304 (2005), no. 1, 158–169.
Consequently, a map $T \colon U \to V$ contracts the inner product if and only if $T = \alpha S$, where S is an isometry and $|\alpha| \leq 1$.
Edit: Here's a simple proof of the assertion that an orthogonality-preserving linear map between finite-dimensional inner product spaces is a scalar multiple of an isometry. Let $T \colon U \to V$ be such a map and fix an orthonormal basis $\{e_1, \ldots, e_n\}$ for $U$. Observe that $e_i + e_j \perp e_i - e_j$. Thus
$$ 0 = \langle T(e_i + e_j), T(e_i - e_j) \rangle = \langle Te_i, Te_i \rangle - \langle Te_j, Te_j \rangle. $$
So we may set $\alpha = \langle Te_i, Te_i \rangle$; this is a nonnegative constant independent of $i$. In particular, if $T$ kills one $e_i$, it kills all the others. It follows that either $T=0$ or else $\{Te_1, \ldots, Te_n\}$ is an orthogonal basis for the range of $T$. In the latter situation, an easy computation yields
$$ \|Tx\|^2 = \sum_i \frac{\langle frac{|\langle Tx, Te_i \rangle^2}{\langle rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i \frac{\langle frac{|\langle \sum_j \langle x, e_j \rangle Te_j, Te_i \rangle^2}{\langle rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i \langle |\langle x,e_i \rangle^2 rangle|^2 \langle Te_i, Te_i \rangle = \alpha \|x\|^2 $$
for all $x \in U$. It follows that $\frac{1}{\sqrt{\alpha}}T$ is an isometry.
