$A\cong A^{op}$ means $A$ is commutative Sorry for my bad English.
Let $A$ be (not necessarily commutative) ring.
By here, if $A\cong A^{op}$ as ring,  $A$ is commutative, but I cannot proof.
Is this trivial? Please help me.
 A: If $A$ is commutative then trivially $A\cong A^{op}$.
I think that the converse is not true, and I think that the following is a counter example:
Let $\mathbb H$ be the Quaternion algebra. Define
$$
f(i)=i \\
f(j)=k \\
f(k)=j  \\
$$
Then $f$ extends to an isomorphism $f: \mathbb H \to \mathbb H^{opp}$.
A: [This just expands upon my comment, since the answer to the question I asked provides another solution]
Given any ring $A$, the map $$(A\times A^{op})^{op} \to A\times A^{op} : (x,y) \mapsto (y,x)$$
will be a ring isomorphism.  If $A$ were non-commutative, then $B = A \times A^{op}$ is non-commutative but $B^{op} \cong B$.
In particular, if your definition of a ring doesn't require a multiplicative identity, then this is succinctly stated as "Every ring $A$ naturally injects into a ring $B$ such that $B^{op} \cong B$"
A: Indeed, every division algebra that has order two in the Brauer group of its center (such as the quaternions) is a counterexample.
But surely the simplest to grasp counterexample is the matrix ring $R=M_{2\times 2}(\Bbb{R})$. It is not commutative. And taking the transpose, i.e. the mapping
$$A\mapsto A^T$$
is an isomorphism from $R$ to $R^{opp}$ by the known rule $(AB)^T=B^TA^T$.
