PeterDonis said:
Bras are defined by the scalars that they map kets to. It's impossible to have multiple bras that all map the same ket to the same scalar. So there can only be one bra that maps a given ket to any scalar, including the scalar that is that ket's squared norm.
You can prove it directly from the Hilbert-space axioms. Technically one must be a bit careful, because everything I say refers to bound linear forms, i.e., those that are continuous in the sense of the norm of the Hilbert space.
So let ##L: H \rightarrow \mathbb{C}## be such a linear form. Further the separable Hilbert space (there is only one up to isomorphism) has a complete orthonormal set ##|u_n \rangle## (##n \in \mathbb{N}##), ##\langle u_j|u_k \rangle=\delta_{jk}##, i.e., for each vector ##|\psi \rangle## there's a unique series ##\psi_n## such that
$$|\psi \rangle=\sum_{n=1}^{\infty} u_n |\psi_n \rangle$$
and
$$\psi_n =\langle u_n|\psi \rangle.$$
Now one can define another series ##L_n## via
$$L_n^*=L(|u_n \rangle).$$
Since ##L## is continuous for all ##|\psi \rangle## you then have
$$L(|\psi \rangle)=L(\sum_{n=1}^{\infty} \psi_n |u_n \rangle=\sum_{n=1}^{\infty} \psi_n L(u_n \rangle)=\sum_{n=1}^{\infty} \psi_n L_n^*.$$
Now we define
$$|L \rangle=\sum_{n=1}^{\infty} L_n |u_n \rangle.$$
Then we have
$$\langle u_n|L \rangle=\langle L|u_n \rangle^*=L_n.$$
From this we get, using the completeness of the ##|u_n \rangle##,
$$\langle L|\psi \rangle=\sum_{n=1}^{\infty} \langle L|u_n \rangle \langle u_n \psi \rangle=\sum_n L_n^* \psi_n=L(|\psi \rangle).$$
This proves the existence of a vector ##|L \rangle## such that ##\langle L|\psi \rangle=L(|\psi \rangle)##.
For the uniqueness assume that also ##|L' \rangle## has this property, but then for all vectors ##|\psi \rangle##
$$\langle L' |\psi \rangle-\langle L|\psi \rangle=L(|\psi \rangle)-L(|\psi \rangle)=0.$$
Thus ##|L'-L \rangle## is a vector for which ##\langle L'-L|\psi \rangle=0## for all ##\psi## and thus also for ##|\psi \rangle=|L \rangle-|L'\rangle##. So we find
$$\langle L-L'|L-L' \rangle=\|L-L'\|^2=0 \; \Rightarrow \; |L \rangle=|L' \rangle,$$
i.e., ##|L \rangle## is the unique vector with the property that ##L(|\psi \rangle)=\langle L|\psi \rangle## for all ##|\psi \rangle \in H##.