Dear Daniel, the answer is yes.

An easy key lemma:

Lemma: Let $\alpha: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ be a continuous character, and $n \geq 1$ an integer. Then there exists one and only one character $\beta: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \alpha$.

Proof: Using the isom $\mathbb R^\ast_+ \simeq \mathbb R$, and the polar decomposition, we are reduce to prove that if $\alpha: \mathbb R \rightarrow \mathbb R$ (resp. $\alpha: \mathbb R \rightarrow \mathbb R / \mathbb Z$) then there exists a unique $\beta$ of the same type such that $\alpha=n \beta$. But any character $\alpha$ as above is of the form $\alpha(x)=ax$ (resp. $\alpha(x)= ax \pmod{\mathbb Z}$) for a unique $a \in \mathbb R$. It is thus clear that taking $\beta(x) = (a/n) x$ works and is the unique possible choice for $\beta$. QED

Nor Consider the composed map: $i: \mathbb R^\ast_+ \hookrightarrow I_k^\infty \hookrightarrow I_k \rightarrow I_k/k^\ast$, where the first map is the diagonal embedding of $\mathbb R^\ast$ in each of the component at infinity of $I_k$ (the product f which I call $I_k^\infty$). This map is clearly injective. Now call a Hecke character $\chi: I_k/k^\ast \rightarrow \mathbb C^\ast$ good if it is trivial on the image of $i$.

Prop: for any Hecke character $\chi$, there are exactly one good character in its equivalence class.

Proof: If $\chi$ and $\chi'$ are good characters in the same class, then $\psi = \chi (\chi')^{-1}$ is trivial on $I_k^1$ and $i(\mathbb R^\ast)$, hence on $I_k$ since any element $x$ on $I_k$ can be written as $(x/ i(y)) i(y)$, with $y = |x|^{1/n} \in \mathbb R^\ast_+$, hence is in $I_k^1 i(\mathbb R_+^\ast)$. Hence the uniqueness. For the existence, by the lemma there exists $\beta: \mathbb R_\ast^+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \chi \circ i$, and consider $\chi' = \chi\ \ \beta^{-1}(|\bullet|)$.

Dear Daniel, the answer is yes.

An easy key lemma:

Lemma: Let $\alpha: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ be a continuous character, and $n \geq 1$ an integer. Then there exists one and only one character $\beta: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \alpha$.

Proof: Using the isom $\mathbb R^\ast_+ \simeq \mathbb R$, and the polar decomposition, we are reduce to prove that if $\alpha: \mathbb R \rightarrow \mathbb R$ (resp. $\alpha: \mathbb R \rightarrow \mathbb R / \mathbb Z$) then there exists a unique $\beta$ of the same type such that $\alpha=n \beta$. But any character $\alpha$ as above is of the form $\alpha(x)=ax$ (resp. $\alpha(x)= ax \pmod{\mathbb Z}$) for a unique $a \in \mathbb R$. It is thus clear that taking $\beta(x) = (a/n) x$ works and is the unique possible choice for $\beta$. QED

Now consider the composed map: $i: \mathbb R^\ast_+ \hookrightarrow I_k^\infty \hookrightarrow I_k \rightarrow I_k/k^\ast$, where the first map is the diagonal embedding of $\mathbb R^\ast$ in each of the component at infinity of $I_k$ (the product of which I call $I_k^\infty$). This map is clearly injective. Now call a Hecke character $\chi: I_k/k^\ast \rightarrow \mathbb C^\ast$ good if it is trivial on the image of $i$.

Prop: for any Hecke character $\chi$, there are exactly one good character in its equivalence class.

Proof: Note that $|i(x)|=x^n$ if $n=[k:\mathbb Q]$ (recall that the $|\ |$ on the component $\mathbb C$ is the square of the complex modulus).

If $\chi$ and $\chi'$ are good characters in the same class, then $\psi = \chi (\chi')^{-1}$ is trivial on $I_k^1$ and on $i(\mathbb R^\ast)$, hence on $I_k$ since any element $x$ on $I_k$ can be written as $(x/ i(y)) i(y)$, with $y = |x|^{1/n} \in \mathbb R^\ast_+$, hence is in $I_k^1 i(\mathbb R_+^\ast)$. Hence the uniqueness. 

For the existence, by the lemma there exists $\beta: \mathbb R_\ast^+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \chi \circ i$, and consider $\chi' = \chi\ \ \beta^{-1}(|\bullet|)$.