Even in finite dimensions you can easily change the inner product. Let $\{ e_{n} \}$ be an orthonormal basis of $\mathbb{C}^{N}$ and define the new inner product
$$
(x,y)_{\mbox{new}}=\sum_{n=1}^{N}\lambda_{n}(x,e_{n})(e_{n},y).
$$
where $\lambda_{n} > 0$ for all $n$.
All the norms are equivalent on $\mathbb{C}^{N}$. This can be written as
$$
(x,y)_{\mbox{new}}=(Ax,y)_{\mbox{old}}
$$
where $A$ is a positive definite selfadjoint matrix. In finite dimensions, this describes every possible inner-product. The inner products are in one-to-one correspondence with positive definite matrices. Because selfadjoint $A$ can be diagonalized by an orthonormal basis of eigenvectors, $(Ax,y) = \sum_{n}\lambda_{n}x_{n}y_{n}^{\star}$ always looks like a weighted inner product when viewed with respect to a correctly chosen orthonormal basis.
If $X$ is an infinite dimensional linear space on which two topologically equivalent Hilbert inner products are defined, say $(\cdot,\cdot)$ and $(\cdot,\cdot)_{1}$, then the same thing happens. There exists a unique positive bounded selfadjoint $A$ such that
$$
(x,y)_{1}=(Ax,y),\;\;\; x,y\in X.
$$
But it also goes the other way: $(x,y)=(Bx,y)_{1}$ where $B$ is positive. You end up with $(x,y)=(Bx,y)_{1}=(ABx,y)$ which gives $AB=I$. Similarly $BA=I$. The existence of such $A$ and $B$ comes from the Lax-Milgram Theorem, which is proved using the Riesz Representation Theorem for bounded linear functionals on a Hilbert Space.
But there are bounded positive linear operators $A$ on a Hilbert space $X$ which are not positive definite. Such an $A$ gives rise to $(x,y)_{1}=(Ax,y)$ with $\|x\|_{1} \le C\|x\|$ for a constant $C$, but the reverse inequality need not hold, which can lead to an incomplete $X$ under $\|\cdot\|_{1}$. For example, let $X=L^{2}[0,1]$ with the usual inner product. Define a new inner product by $(f,g)_{1}=\int_{0}^{1}xf(x)g(x)\,dx$. This is achieved as $(Af,g)$ where $Af=xf(x)$. This space is not complete because $1/\sqrt{x}$ is in the completion of $L^{2}$ under the norm $\|\cdot\|_{1}$. The completion of $(X,\|\cdot\|_{1})$ consists of $\frac{1}{\sqrt{x}}L^{2}[0,1]$. However, if you instead define $\|f\|_{1}^{2}=\int_{0}^{1}(x+\epsilon)|f(x)|^{2}\,dx$ for some $\epsilon > 0$, then you end up with an equivalent norm on $X$.
The Riemannian metric provides a natural identification of the cotangent space with the tangent space. Therefore the cotangent space is also naturally equipped with a Euclidean (or if you prefer Hilbert-space) metric. The simplest way of thinking of the identification is to send a tangent vector $v$ to the covector given by the inner product with $r$, namely $\langle v,\cdot\rangle$.
So they are naturally isomorphic as Hilbert spaces, Banach spaces, and metric spaces.
Expressed in terms of indices, the relation is simply this. If $v_i$ represents a vector and $\alpha^j$ a covector then the relation $v_i=g_{ij}\alpha^j$ (summation over a repeated index as usual) gives a way of passing between a vector and its corresponding covector. Here $g_{ij}$ is the Riemannian metric.
Best Answer
There are two notions of what it means for a(n infinite dimensional) manifold to have a Riemannian structure. A strong Riemannian structure means a (smooth) choice of inner product on each tangent space which induces an isomorphism of each tangent space with its corresponding cotangent space. A weak Riemannian structure simply means a (smooth) choice of inner product on each tangent space.
Strong Riemannian structures only exist if the manifold is modelled on a Hilbert space, and even then they have to be chosen correctly (so the usual $L^2$-metric on the space of $L^{1,2}$-loops is not a strong structure, even though the manifold is Hilbertian). Weak Riemannian structures exist much more widely. For a weak Riemannian structure you only need to know that the manifold admits smooth partitions of unity and that the model spaces admit continuous inner products. So, for example, continuous loops in a smooth manifold admit a weak Riemannian structure but not a strong one.
Although strong Riemannian structures are very good for generalising much of ordinary differential geometry to infinite dimensions, there are occasions where the requirement of having a Hilbert manifold is too strong, and one can get away with merely having a weak Riemannian structure. I've written an article where having a weak Riemannian structure on the space of smooth loops was an essential step and where the construction would not have worked on a Hilbertian manifold (though actually it was a co-Riemannian structure that I needed).
(Declaration of interests: I've actually proposed a refinement of the "weak/strong" classification as I found it too harsh. See my article here for this, and the above-mentioned result, and a load of examples of spaces with different types of Riemannian structure.)