Galois descent for a semisimple automorphism

Question

Let $K$ be a perfect field and $\overline{K}$ be the algebraic/separable closure. Let $V$ be a finite dimensional $K$-vector space, and let $V_{\overline{K}} = V \otimes_K \overline{K}$. Given an endomorphism $S:V \to V$, define $S_{\overline K}:V_{\overline K} \to V_{\overline K}$ as $S \otimes 1$.

As $K$ is perfect, $\overline K/K$ is Galois, so let $G = \operatorname{Gal}(\overline{K} /K)$. For $\sigma \in G$ and $\overline{S} \in \operatorname{Aut}_{\overline K}(V_{\overline K})$, let $$ \overline{S}^{\sigma} = (1 \otimes \sigma) \circ \overline{S} \circ (1 \otimes \sigma^{-1}) $$ This defines an action of $G$ on $\operatorname{Aut}_{\overline K}(V_{\overline K})$. It is clear that if $S \in \operatorname{Aut}_K(V)$, then $S_{\overline K}$ is fixed by this action, that is, $S_{\overline K} \in \operatorname{Aut}_{\overline K}(V_{\overline K})^G$.

I seek a partial converse to the previous statement. The full converse would be, "If $\overline{S} \in \operatorname{Aut}_{\overline K}(V_{\overline K})^G$, then there exists $S \in \operatorname{Aut}_K(V)$ such that $\overline{S} = S_{\overline K}$." I'm not sure if it is true, but I would be happy if I could prove this with the additional hypothesis that $\overline S$ is semisimple (semisimple = diagonalizable = $\exists$ a basis of $V_{\overline K}$ of eigenvectors for $\overline S$).

Partial results/progress:

1. I can prove it in the case where $\overline{S}$ acts by a scalar; this basically comes down to Galois theory.

2. Supposing $\overline{S}$ is fixed by the Galois action, if $\lambda$ is an eigenvalue for $\overline S$ with eigenvector $w$, and $\sigma \in G$, then $(1 \otimes \sigma)(w)$ is also an eigenvector with eigevalue $\sigma(\lambda)$. This is immediate from $\overline S = \overline S^{\sigma}$.

3. One thing that initially seemed promising is that to show $\overline S$ comes from $V$, it would suffice to show that all the eigenvalues of $\overline S$ are in $K$. However, it may happen that $\overline S$ is semisimple yet has eigenvalues not lying in $K$. I have very concrete example of this involving $K = \mathbb{R}$ where $\overline S$ is just a $2 \times 2$ rotation matrix, which I can provide details on if necessary. Suffice to say, this is not a way to prove the claim, because it would require proving something that isn't true.

4. Since $\overline{S}$ has only finitely many eigenvalues $\lambda_1, \ldots, \lambda_n$, rather than deal with $\overline K$, which may be infinite dimensional over $K$, we can just consider $\overline S$ as defined over the finite extension $K(\lambda_1, \ldots, \lambda_n)$ of $K$. This may or may not help.