Is $x^\ast$ a critical point when $\sum_{\lambda_i=1}^k\lambda_i$ and ${\rm D}f(x^\ast)^\ast\lambda=\sum_{i=1}^k\lambda_i\nabla f_i(x^\ast)=0$?

Question

Let $k,d\in\mathbb N$ and $f:\mathbb R^d\to\mathbb R^k$ be Fréchet differentiable. Say that $x^\ast\in\mathbb R^d$ is critical if $$\not\exists x\in\mathbb R^d:f(x)<f(x^\ast)\tag1$$ (component-wisely). It's easy to show that $x^\ast\in\mathbb R^d$ is critical if and only if there is a $\lambda\in[0,\infty)^k$ with $\sum_{i=1}^k\lambda_i=1$ and $${\rm D}f(x^\ast)^\ast\lambda=\sum_{i=1}^k\lambda_i\nabla f_i(x^\ast)=0\tag2.$$

Question 1: Now, let $x^\ast\in\mathbb R^d$ and $\lambda\in\mathbb R^k$ with $(2)$. Does $(1)$ it still need to hold?

Question 2: And how precisely is $(1)$ related to the linear independence of $\nabla f_1(x^\ast),\ldots,\nabla f_k(x^\ast)$?

Regarding the first question: Clearly, the only thing missing is the nonnegativity and normalization of the weights, but I guess this could be crucial.

Regarding the second question: I know that $\nabla f_1(x^\ast),\ldots,\nabla f_k(x^\ast)$ are the rows of $A:={\rm D}f(x)\in\mathbb R^{k\times d}$ and that they are linearly independent if and only if one of the following equivalent conditions hold:

1. $\forall\lambda\in\mathbb R^k:A^\ast\lambda=0\Rightarrow\lambda=0$;
2. $A^\ast$ is injective;
3. $A$ is surjective.

However, can it be inferred from the linear independence of $\nabla f_1(x^\ast),\ldots,\nabla f_k(x^\ast)$ that $x^\ast$ is critical?