Let $K$ be a field and $G$ be an algebraic group. Specifically $O(n)$ or $Sp\_{2n}$. Is it true that for any ring $A$ over $K$ , $G(A)\cong G(A[x])$. 
Is there any reference for such kind of results?



$A[x]$ points of an algebraic group
The comments show that the answer to the question is negative if $G$ contains a copy of $\mathbb G\_a$. The examples $O(n)$ and $Sp(n)$ in the question suggest, though, that the emphasis is on anisotropic $G$. It is interesting that in this case the answer is affirmative to some extent. 


Let, e.g., $K=\mathbb R$, let $G$ be a linear algebraic group not containing $\mathbb G\_a$. This means that $G^0=SC$ where $S$ is a split torus and $C$ is anisotropic, i.e., $C(\mathbb R)$ is compact. Claim: Under these conditions $G(K)\overset\sim\to G(K[x])$.


An element of $G(K[x])$ is a morphism $f:\mathbb A^1\to G$. If $G=S$ then $f$ is constant since $K[x]^\times=K^\times$. If $G=C$ then the image of $f$ is closed. The reason for this is that $C$ is affine and therefore $f$ can't extend to $\mathbb P^1$. But, since $C(\mathbb R)$ is compact, the image of $\mathbf A^1(\mathbb R)$ in $C(\mathbb R)$ can't be closed unless $f$ is constant. The mixed case follows from these two subcases.


From this it follows easily that $G(A)\overset\sim\to G(A[x])$ when $A=\mathcal O(X)$ where $X$ is an irreducible normal affine $K$-variety such that $X(K)$ is Zariski dense in $X$.


With some effort, the argument can be extended to any field of characteristic zero.


