Suppose $H$ is a reproducing kernel Hilbert space and $K_{1}\left(x,\cdot\right)$ and $K_{2}\left(x,\cdot\right)$ two reproducing kernels with respect to two equivalent inner products on this space. What can we say about these kernels in general. Are they equal? Is there any theory on this topic?

1$\begingroup$ Of course, they are not equal: If $\langle f,g\rangle_2=\lambda \langle f,g\rangle_1$ with $\lambda>0$ you get from $f(x)= \langle f,K_i(x,\cdot)\rangle_i$ that $\lambda K_2 = K_1$. $\endgroup$ – Jochen Wengenroth Feb 24 '14 at 12:21
As shown in the comments, the answer to your first question is no. However, there is a relationshp between the two reproducing kernels which might answer your second question. If you have two equivalent Hilbert norms on a space, then they are related by a positive, invertible operator $A$ in the sense that $$(fg)_2=(fAg)_1.$$ The relationship between the kernel functions is then as follows. By the definition, $K_1(x,y)$ belongs to the Hilbert space as a function of $y$ for each $x$. We then define $K_2(x,y)$ to be the image of this element under $A$.
The most transparent situation is where the Hilbert space is $L^2$ and $A$ is multiplication by a suitable measurable function (positive, bounded and bounded away from zero). The second norm is then a weighted $L^2$ norm. Indeed, this is the only case in a certain senseby the spectral theorem in the version that every self adjoint operator is unitarily equivalent to a multiplication operator.