To expand on my comment to the question, we have the following algebraic construction (I think originally due to Robert Brown, 'A characterization of spin representations'):

Let $V$ be a quadratic space over a field $k$ of characteristic $\neq 2$. Attached to this is the Clifford algebra $C=C(V)$: it is equipped with a $\mathbb{Z}/2\mathbb{Z}$-grading $C=C^+\oplus C^-$ and an embedding $V\hookrightarrow C^-$. The general Spin group $GSpin(V)$ consists of units in $C^+$ that preserve $V$ under conjugation. The group $Spin(V)$ is the sub-group of elements that have trivial spinor norm. So, to describe $Spin(V)$ as an automorphism group, it suffices to do so for $GSpin(V)$.

Let $H$ be the graded vector space $C$ viewed as a representation of $GSpin(V)$ via left multiplication: it is also a right $C$-module via right multiplication. Then $GSpin(V)$ clearly lies within the group $U(H)$ of $C$-equivariant, grading preserving automorphisms of $H$.

Set $E=End(H)$: this is a representation of $GSpin(V)$ via conjugation. Define a bilinear form $\{,\}:E\times E\to k$ by $$\{f,g\}=\frac{1}{2^{dim(V)}}trace(fg).$$

Now choose a basis $\{v_i\}$ for $V$, and let $A=(v_i\cdot v_j)_{i,j}$ be the inner product matrix attached to this basis. Set $(b_{i,j})=B=A^{-1}$. Define an endomorphism $\pi:E\to E$ by the formula:

$$\pi(f)(h)=\sum_{i,j}b_{i,j}\{v_i,f\}v_jh.$$

Clearly, the image of $\pi$ is $V\subset E$, where $V$ acts on $H$ via left multiplication. Let $G'\subset U(H)$ be the stabilizer of the endomorphism $\pi$. Then $G'$ preserves $V$ via conjugation, and is therefore contained in $GSpin(V)$. On the other hand, it is not hard to see that $GSpin(V)$ stabilizes $\pi$.

So we see that $GSpin(V)$ can be described as the group of $C$-equivariant, grading preserving automorphisms of $H$ that also stabilize $\pi$.

allautomorphisms, not merely to be a subgroup of an automorphism group. So the examples with $L^2$-functions don't fit that situation (e.g., for similar reasons, the fact that Spin groups are found inside Clifford algebras doesn't immediately furnish an answer when $n$ is not very small). $\endgroup$1more comment