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Map on Picard groups has a kernel.
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Let $K = \mathbb R$, and let $C = \{x^2+y^2+z^2 = 0\} \subseteq \mathbb P^2_K$ be the conic without a rational point. The projection $\pi \colon C_{\bar K} \to C$ induces a pullback map $$\pi^* \colon \operatorname{Pic}(C) \to \operatorname{Pic}(C_{\bar K}) = \mathbb Z$$ identifyingmapping $\operatorname{Pic}(C)$ withto the subgroup $2\mathbb Z \subseteq \mathbb Z$. In particular, there is no degree $1$ line bundle on $C$ of degree $1$.

Now equip the vector bundle $\mathcal O(1) \oplus \mathcal O(1)$ on $C_{\bar K}$ with a Galois action by interchanging the factors. This descends to a bundle $\mathcal E$ on $C$, which can equivalently be described as $\pi_* \mathcal O(1)$.

Claim. The rank $2$ vector bundle $\mathcal E$ is stable on $C$, but its pullback to $C_{\bar K}$ is not.

Proof. The second statement is clear. For the first, note that $\mathcal E$ is semistable since it is so over $\bar K$. Thus, it suffices to show that it does not have anycontain a line bundle of degree $1$ subbundles. But $C$ does not have any line bundle of degree $1$. $\square$

Let $K = \mathbb R$, and let $C = \{x^2+y^2+z^2 = 0\} \subseteq \mathbb P^2_K$ be the conic without a rational point. The projection $\pi \colon C_{\bar K} \to C$ induces a pullback map $$\pi^* \colon \operatorname{Pic}(C) \to \operatorname{Pic}(C_{\bar K}) = \mathbb Z$$ identifying $\operatorname{Pic}(C)$ with the subgroup $2\mathbb Z \subseteq \mathbb Z$. In particular, there is no degree $1$ line bundle on $C$.

Now equip the vector bundle $\mathcal O(1) \oplus \mathcal O(1)$ on $C_{\bar K}$ with a Galois action by interchanging the factors. This descends to a bundle $\mathcal E$ on $C$, which can equivalently be described as $\pi_* \mathcal O(1)$.

Claim. The rank $2$ vector bundle $\mathcal E$ is stable on $C$, but its pullback to $C_{\bar K}$ is not.

Proof. The second statement is clear. For the first, note that $\mathcal E$ is semistable since it is so over $\bar K$. Thus, it suffices to show that it does not have any degree $1$ subbundles. But $C$ does not have any line bundle of degree $1$. $\square$

Let $K = \mathbb R$, and let $C = \{x^2+y^2+z^2 = 0\} \subseteq \mathbb P^2_K$ be the conic without a rational point. The projection $\pi \colon C_{\bar K} \to C$ induces a pullback map $$\pi^* \colon \operatorname{Pic}(C) \to \operatorname{Pic}(C_{\bar K}) = \mathbb Z$$ mapping $\operatorname{Pic}(C)$ to the subgroup $2\mathbb Z \subseteq \mathbb Z$. In particular, there is no line bundle on $C$ of degree $1$.

Now equip the vector bundle $\mathcal O(1) \oplus \mathcal O(1)$ on $C_{\bar K}$ with a Galois action by interchanging the factors. This descends to a bundle $\mathcal E$ on $C$, which can equivalently be described as $\pi_* \mathcal O(1)$.

Claim. The rank $2$ vector bundle $\mathcal E$ is stable on $C$, but its pullback to $C_{\bar K}$ is not.

Proof. The second statement is clear. For the first, note that $\mathcal E$ is semistable since it is so over $\bar K$. Thus, it suffices to show that it does not contain a line bundle of degree $1$. But $C$ does not have any line bundle of degree $1$. $\square$

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Let $K = \mathbb R$, and let $C = \{x^2+y^2+z^2 = 0\} \subseteq \mathbb P^2_K$ be the conic without a rational point. The projection $\pi \colon C_{\bar K} \to C$ induces a pullback map $$\pi^* \colon \operatorname{Pic}(C) \to \operatorname{Pic}(C_{\bar K}) = \mathbb Z$$ identifying $\operatorname{Pic}(C)$ with the subgroup $2\mathbb Z \subseteq \mathbb Z$. In particular, there is no degree $1$ line bundle on $C$.

Now equip the vector bundle $\mathcal O(1) \oplus \mathcal O(1)$ on $C_{\bar K}$ with a Galois action by interchanging the factors. This descends to a bundle $\mathcal E$ on $C$, which can equivalently be described as $\pi_* \mathcal O(1)$.

Claim. The rank $2$ vector bundle $\mathcal E$ is stable on $C$, but its pullback to $C_{\bar K}$ is not.

Proof. The second statement is clear. For the first, note that $\mathcal E$ is semistable since it is so over $\bar K$. Thus, it suffices to show that it does not have any degree $1$ subbundles. But $C$ does not have any line bundle of degree $1$. $\square$