= Solution
The finite passive <dilatation> is $x'=e^{-\alpha}x$, $\phi'(x')=e^\alpha\phi(x)$. To compare at the same argument, invert the coordinate change:
$$
\phi'(x)=e^\alpha\phi(e^\alpha x)=\phi(x)+\alpha\phi(x)+\alpha x^\mu\partial_\mu\phi(x)+O(\alpha^2).
$$
Consequently the fixed-argument infinitesimal <variation> is
$$
\boxed{\bar\delta\phi(x)=\alpha(\phi+x^\mu\partial_\mu\phi).}
$$
This also follows from $\delta\phi=\alpha\phi$ and $\delta x^\mu=-\alpha x^\mu$. The plus sign on the derivative term is a consequence of this specified contraction of coordinates.
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