Convex combination of cyclically monotone sets

Question

I want to show the following statement, but I am not sure how.

Proposition(?): Let $C \in \mathbb{R}^d$ be a compact convex set, and let $u, v : C \to \mathbb{R}$ be smooth convex functions. Suppose $$ \{\nabla u (x) \mid x \in C\} = \{\nabla v (x) \mid x \in C\} =: D \subset \mathbb{R}^d , $$ and $D$ is convex. Then, for $\lambda \in (0, 1),$ $$ \{(1 - \lambda) \nabla u (x) + \lambda \nabla v (x) \mid x \in C\} = D $$ holds.

I know:

1. Since $u, v$ are convex, each of $\nabla u, \nabla v$ is a homeomorphism between $C$ and $D.$
2. Since $(1 - \lambda) u + \lambda v$ is convex, its gradient is also a homeomorphism onto its image.
3. Since $D$ is convex, $\{(1 - \lambda) \nabla u (x) + \lambda \nabla v (x) \mid x \in C\} \subset D$ is clear.
4. The claim for $d = 1$ is not hard to show because $\nabla u, \nabla v$ are increasing functions on a bounded closed interval $C$ in this case.

Is this statement true in general? If so, how can I prove it? In particular, I am not sure how I should use the cyclical monotonicity. (Without it, the statement fails even when $d = 1.$)