I asked this basic question in MSE and got a comment "This belongs to Mathoverflow", so I ask my question here.

Let $G$ be a linear algebraic group over a field $k$, and $H\subset G$ be a $k$-subgroup. Let $K/k$ be a field extension. I need an isomorphism between $G_K/H_K$ and $(G/H)_K$.

In more detail, let $G$ be a linear algebraic group over a field $k$ (of arbitrary characteristic). Let $H\subset G$ be an algebraic $k$-subgroup. Set $X=G/H$, it exists (because $G$ is linear), and it is a $k$-variety (a reduced scheme of finite type over $k$, not necessarily connected). It has a base point $x_0=1\cdot H\subset X(k)$. The group $G$ acts on $X$ by $$(g',gH)\mapsto g'gH,$$ and the scheme-theoretic stabilizer of $x_0$ in $G$ is $H$.

Now let $K/k$ be a field extension (not necessarily finite). The base change $G_K:=G\times_k K$ acts on $X_K$, and the scheme-theoretic stabilizer of $x_0\in X(k)\subset X(K)=X_K(K)$ in $G_K$ is $H_K$. Consider the map (morphism) $$\lambda\colon G_K\to X_K,\quad g\mapsto g\cdot x_0\,.$$ Clearly this map is constant on the orbits of $H_K$ acting on the right on $G_K$, hence it induces a morphism $$\bar\lambda\colon G_K/H_K\to X_K\,,$$ see Borel, Linear Algebraic Groups, 2nd edition, Section 6.

Question. How can I prove that $\bar\lambda$ is an isomorphism of $K$-varieties?

I would be grateful for a reference or proof.