Are (EC) DSA blind signatures possible?
I am not aware of any existing scheme allowing this easily (excepted for PureEdDSA maybe, but I wouldn't classify it as an ECDSA variant).

However I do not believe this is impossible. So let's try to do this for ECDSA:

We have a message $m$ whose hashed value $H(m)$ is converted into an integer $z$ which gets signed into the signature $(r,s)$ by the signer.

Now the verifier will basically rely on
$$\begin{aligned}
C&amp;=u_1\times G+u_2\times Q\\
&amp;=u_1 \times G +u_2d \times G\\
&amp;=(u_1+u_2 d)\times G \\
&amp;=(zs^{-1}+rds^{-1})\times G \\
&amp;=(z+rds)\color{red}{s^{-1}}\times G \\
&amp;=(z+rd)\color{red}{(z+rd)^{-1}(k^{-1})^{-1}}\times G \\
&amp;=k\times G 
\end{aligned}$$
And will say that it verifies iff $C_x==r$
And as you can see, if you can change $r,s,z$ then you can force a coefficient in the latest line to cancel "almost out", by taking $bs,abr,abz$ instead and would end up with $$(abz+abrd)\color{red}{b^{-1}(z+rd)^{-1}(k^{-1})^{-1}}\times G \\
= ak\times G $$

Now the difficult part is to find out if it is possible to get $a,b,m'$ such that $H(m')$, once converted into $z'$ is such that $z'=abz$ and such that $(ak\times G)_x \equiv abr\mod n$...

I do not see why this couldn't be possible, being given so much slack on the variables... But since I don't have time right now to actually try it out, I'll try to do it later or another day and will edit this answer accordingly.