1. Let the magnitudes of the vectors be denoted as usual, and use the given condition $|\vec x+\vec y|=|\vec x|.$

2. Squaring both sides: $|\vec x+\vec y|^2=|\vec x|^2.$ Using $|\vec x+\vec y|^2=|\vec x|^2+|\vec y|^2+2\vec x\cdot \vec y,$ we get $|\vec x|^2+|\vec y|^2+2\vec x\cdot \vec y=|\vec x|^2.$ Hence, $|\vec y|^2+2\vec x\cdot \vec y=0$ or $2\vec x\cdot \vec y=-|\vec y|^2.$

3. Now use the second condition: $2\vec x+\lambda \vec y$ is perpendicular to $\vec y$. Therefore, $(2\vec x+\lambda \vec y)\cdot \vec y=0.$ Expanding: $2\vec x\cdot \vec y+\lambda |\vec y|^2=0.$

4. Substitute from step 2: $- |\vec y|^2+\lambda |\vec y|^2=0.$ $|\vec y|^2(\lambda-1)=0.$

5. Since $\vec y$ is a non-zero vector, $|\vec y|^2\neq 0$. Hence, $\lambda-1=0\Rightarrow \lambda=1.$

Therefore, the required integer is $\boxed{1}.$