A question on Haar measure on local field. Let $F$ be a local field of characteristic 0, and $f:F\rightarrow \mathbb{C}$ be an integrable function. Is the following formulation valid?
$
\int_{F^\times}f(x^2) d^\times x=\int_{F^{\times 2}}f(x) d^\times x
$
where $d^\times x$ is a chosen multiplicative measure on $F^\times$.
It seems if $F=\mathbb{R}$, this is true. But is it true for any local field?
 A: This is false for non-archimedean local fields. On the other hand, as Paul Garrett explained, the two sides are equal up to an absolute constant.
Let $F=\mathbb{Q}_2$, and let $f$ be the characteristic function of $1+8\mathbb{Z}_2$. Then the left hand side is the measure of $1+2\mathbb{Z}_2$, while the right hand side is the measure of $1+8\mathbb{Z}_2$, hence they differ by a factor of $4$.
Let $F=\mathbb{Q}_p$ for $p>2$, and let $f$ be the characteristic function of $1+p\mathbb{Z}_p$. Then the left hand side is the measure of $\cup(\pm 1+p\mathbb{Z}_p)$, while the right hand side is the measure of $1+p\mathbb{Z}_p$, hence they differ by a factor of $2$.
A: Since the subgroup of squares is open, the restriction of Haar measure from the full multiplicative group to the squares is a Haar measure on the subgroup of squares, and Haar measure is unique up to scalars.
A: I think for an appropriate choice of the measure on $F^{\times 2}$, I still believe this is true, which is essentially what Paul Garret mentioned. Indeed $F^{\times 2}\cong F^{\times}/ \{\pm1\}$, and the function $x↦f(x^2)$ can be viewed as a function on this quotient. Then 
\begin{equation}
\int_{F^\times}f(x^2)d^\times x=\int_{F^{\times}/{\pm 1}}\sum_{\{\pm 1\}}f(y^2)d^\times y
\end{equation}
where $d^\times y$ is the quotient measure. Then by identifying $F^{\times}/{±1}$ with $F^{\times 2}$, one can see that the left hand side is 2 times $\int_{F^{\times 2}}f(y)d^\times y$.
