Dear Daniel, the answer is yes. 

An easy key lemma:

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**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

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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 $\chi' = \chi\ \  \beta^{-1}(|\bullet|)$.