You are essentially asking about the structure of the set $D_X$ of all dual frames $Y$ of a given frame $X$. As mentioned by Dustin G. Mixon, there is always one preferred dual frame, namely the canonical dual frame $X'$. Now regarding uniqueness: $D_X=(X')$ if and only if $X$ is a Riesz basis.
If $H$ has finite dimension, then a frame $X$ is a Riesz basis if and only if $|X|=\dim H$. So you will get infinitely many duals as soon as $|X|\geq \dim H+1$, and only the canonical dual when $|X|=\dim H$.
In general, the set $D_X$ has a natural structure of affine space via the consideration of idempotents. Here is a detailed outline.
Analysis/frame/synthesis operator: for any family $(x_j , j\in J)$, we consider the so-called analysis operator $\theta_X:H\longrightarrow \mathbb{C}^J$ sending $x$ to $((x,x_j))_{j\in J}$. By definition, $X$ is a frame if $\theta_X$ is bounded above and below from $H$ to $\ell^2(J)$. That is, $\theta_X$ is bounded and the positive so-called frame operator $\theta_X^*\theta_X$ is invertible (with spectrum in $[A,B]$, the constants of the definition). Denoting by $(e_j,j\in J)$ the canonical basis of $\ell^2(J)$, note that the adjoint (called the synthesis operator) satisfies $\theta_X^*e_j=x_j$.
A useful projection: when $X$ is a frame, the projection $p_X$ onto the range of $\theta_X$ is given by $p_X=\theta_X(\theta_X^*\theta_X)^{-1}\theta_X^*$.
Riesz basis: a frame $X$ is a Riesz basis if and only if $\theta_X^*$ is injective, if and only if $\theta_X$ is surjective, if and only if $\theta_X$ is invertible ($\theta_X^*$ yielding the equivalence between $X$ and the canonical basis of $\ell^2(J)$). That's equivalent to $p_X=1$.
Dual frame: a frame $Y$ is called a dual frame of a frame $X$ if $\theta_Y^*\theta_X=\mbox{Id}_H=1$. That is $v=\sum_j (x_j,v)y_j$ for every $v\in H$. That's equivalent to $\theta_X^*\theta_Y=\mbox{Id}_H=1$. That is $v=\sum_j (y_j,v)x_j$ for every $v\in H$.
Canonical dual fame: given a frame $X$, the canonical dual $X'$ of $X$ is given by the formula $\theta_{X'}^*=(\theta_X^*\theta_X)^{-1}\theta_X^*$. It is indeed trivial to check that $\theta_{X'}^*\theta_X=1$.
Idempotent characterization of duality: it is easy to see that, given two frames $X,Y$, they are alternate duals of each other if and only if $\theta_X\theta_Y^*$ is idempotent in $B(\ell^2(J))$.
Affine structure: given a frame $X$, the mapping $Y\longmapsto \theta_X\theta_Y^*$ is a bijection from the set $D_X$ of all duals of $X$ to the set of all idempotents $q\in B(\ell^2(J))$ with range equal to that of $\theta_X$. The latter is simply
$$
D_X\simeq p_X+p_X B(\ell^2(J))(1-p_X)
$$
where $p_X$ corresponds to the canonical dual $X'$.
Since $p_X=0$ is impossible ($\theta_X^*p_X=\theta_X^*\neq 0$), it follows that $D_X$ is a singleton (unique dual) if and only if $p_X=1$ ($X$ Riesz basis). Otherwise, it is of course infinite and the canonical dual is optimal in the sense that it minimizes the norm of the idempotent $\theta_X\theta_Y^*$. It is indeed the only $Y$ for which the latter is $1$.