Are Killing vector fields eigenvectors for extrinsic curvature?

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A quick question I struggle with these days:

Let's consider a 2D smooth surface embedded with $\mathbb{R}^3$. Such a surface features an extrinsic curvature tensor field ($\boldsymbol{\kappa}$) as well as a metric field ($\boldsymbol{\gamma}$).

Let's also assume that this surface has some continuous symmetry and therefore admits a Killing vector field ($\boldsymbol{k}$), i.e. $\exists\: \boldsymbol{k} \: \text{on}\:\mathcal{S}\: |\: \mathcal{L}_{\boldsymbol{k}}(\boldsymbol{\gamma}) = \boldsymbol{0}$, where $\mathcal{L}_a(b)$ stands for the Lie derivative of $b$ along $a$.

At first glance, considering simple surfaces such as cylinders or spheroids, I have the feeling that such a Killing vector field must correspond to an eigenvector field for the extrinsic curvature, i.e. $\boldsymbol{\kappa}\cdot\boldsymbol{k}\propto\boldsymbol{k}$.

My question : Is this actually the case ? If so could someone give me an hint toward the demonstration ? If not, do you have a counter example ?

Thx !