如图所示,已知椭圆\(E:\dfrac{{{y}^{2}}}{{{a}^{2}}}+\dfrac{{{x}^{2}}}{{{b}^{2}}}=1(a > b > 0)\)的长轴长是焦距的\(\sqrt{2}\)倍,短轴右端点为\(A\),\(M(1,0)\)为线段\(OA\)的中点\((\)\(O\)为坐标原点\()\).
\((\)Ⅰ\()\)求椭圆\(E\)的方程;
\((\)Ⅱ\()\)过点\(M\)任作一条直线,与椭圆\(E\)相交于\(P,Q\)两点,试问在\(x\)轴上是否存在定点\(N\),使得\(\overrightarrow{OM}=\overrightarrow{ON}+\lambda (\dfrac{\overrightarrow{NP}}{\left| \overrightarrow{NP} \right|}+\dfrac{\overrightarrow{NQ}}{\left| \overrightarrow{NQ} \right|})(\lambda > 0)\)?若存在,求出点\(N\)的坐标;若不存在,请说明理.