By the same averaging trick that shows that finite-dimensional complex representations of a finite group are unitary with respect to some inner product, your question is equivalent to the one obtained by replacing ambient isotropy groups with linear automorphism groups in the sense of algebraic geometry. Here linear means induced by a linear automorphism of $\mathbf{P}^2$. Actually, for a smooth plane curve of degree $d>3$, all automorphisms are linear, by
H. C. Chang, On plane algebraic curves, Chinese J. Math. 6 (1978), 185-189.
Fix $d>3$. Let $\mathcal{H}_d$ be the moduli space of smooth degree-$d$ curves in $\mathbf{P}^2$, so $\mathcal{H}_d$ is an open subscheme of some projective space. Then there is a stratification of $\mathcal{H}_d$ into finitely many locally closed subschemes such that the automorphism group is constant on each piece. (This could also be stated in terms of the automorphism group scheme of the universal curve over $\mathcal{H}_d$.) In these terms, you are asking for the $0$-dimensional strata, or equivalently the smooth plane curves such that in a punctured neighborhood of the corresponding point of $\mathcal{H}_d$ the automorphism group is strictly smaller.
The analogous question with $\mathcal{H}_d$ replaced by the full moduli space $\mathcal{M}_g$ of curves of genus $g>1$ has been much studied. The direct analogue of your maximally symmetric curves in this setting are the curves said to have "many automorphisms" in Section 3 of the article
Jürgen Wolfart, The obvious part of Belyi's theorem and curves with many automorphisms, pp. 97-112 in: Geometric Galois actions 1, edited by L. Schneps and P. Lochak, LMS Lecture Notes Series 342, Cambridge Univ. Press, 1997.
Wolfart's article contains many references to related work, and mentions some nice theorems. For instance: a smooth projective curve of genus greater than $1$ over $\mathbf{C}$ has many automorphisms if and only if it is a Galois cover of $\mathbf{P}^1$ ramified only above $0,1,\infty$.