A quasigroup is a pair $(Q, \cdot)$ where $Q$ is a non-empty set and $\cdot$ is a binary operation on $Q$ such that for every $(u, v) \in Q^2$ there exists a unique $(x, y) \in Q^2$ such that $u \cdot x = v = y \cdot u$. Let $q$ be an odd prime power, let $\mathbb{F}_q$ denote the finite field of order $q$, and let $\mathcal{R}_q$ denote the set of non-zero squares in $\mathbb{F}_q$. Let $\{a, b\} \subseteq \mathbb{F}_q$ be such that $\{ab, (a-1)(b-1)\} \subseteq \mathcal{R}_q$. Let $\mathcal{Q}_{a, b}$ denote the quadratic quasigroup $(\mathbb{F}_q, *_{a, b})$ where $*_{a, b}$ is defined by \[ \left\{ \begin{array}{ll} x+a(y-x) & \text{if } y-x \in \mathcal{R}_q,\\ x+b(y-x) & \text{otherwise}. \end{array} \right. \] The operation table of a quadratic quasigroup is a quadratic Latin square. Recently, it has been determined exactly when two quadratic quasigroups are isomorphic and the automorphism group of any quadratic quasigroup has been determined. In this paper, we extend these results. We determine exactly when two quadratic quasigroups are isotopic and we determine the autotopism group of any quadratic quasigroup. In the process, we count the number of $2 \times 2$ subsquares in quadratic Latin squares.