As Roy remarked in his answerabove, the delta invariant of the germ of plane curve singularity $f(x,y)=0$ at $p$ is equal to
$\delta(f) = \sum \frac{m_q(m_q-1)}{2}$,
where the sum is extended over all the points $q$ which are "infinitely near" to $p$ and $m_q$ denotes the multiplicity at $q$.
Then $\delta(f)$ is minimal among germs of a given multiplicity when there are no infinitely near points, in other word words when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points.
Now, since the Milnor number is equal to
$\mu(f)=2 \delta(f)-r(f)+1$,
where $r(f)$ is the number of branches, it follows that $\mu(f)$ is minimal among plane singularities of given multiplicity $n$ when $\delta(f)$ is minimal and the number of branches is maximal, in other words when $f=0$ is the "ordinary" $n$-ple point.
As an example, for the ordinary double point (node) $y^2=x^3+x^2$ we have
$(\delta, \mu)=(1,1)$,
whereas the ordinary cusp $y^2=x^3$ satisfies
$(\delta, \mu)=(1,2)$.
Summing up, the ordinary $n$-ple point is the only germ of plane curve singularity which minimizes both $\delta$ and $\mu$, and the corresponding values are
$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.
