$\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}\DeclareMathOperator\GL{GL}$I asked a similar question on math stack exchange here, but I wonder if it may be better received here.

Let $n$ be even, then the standard complex Clifford algebra admits an isomorphism: $$ \mathbb{C}\text{l}(n)\cong \text{End}(\mathbb{C}^N) $$ where $N=2^{n/2}$. We call $\mathbb{C}^N$ the space of Dirac spinors. Let $\gamma:\mathbb{R}^n\rightarrow\text{Cl}(n)$ be the canonical injection, and $f:\text{Cl}(n)\rightarrow \mathbb{C}\text{l}(n)\cong \text{Cl}(n)\otimes_\mathbb{R}\mathbb{C}$ be the inclusion map. The above isomorphism then induces a blinear map called spinor multiplication with a vector: $$ \begin{align*} \mathbb{R}^n\times \mathbb{C}^N&\longrightarrow \mathbb{C}^N\\ (x,\psi)&\longmapsto x\cdot \psi =(f\circ \gamma(x))\cdot \psi \end{align*} $$ where we identify $f\circ \gamma(x)$ with it's unique endomorphism under the aforementioned isomorphism. Identifying $\text{Spin}(n)$ as a subset of the real Clifford algebra then yields a faithful representation of $\text{Spin}(n)$ on $\mathbb{C}^N$, called the spinor representation. We denote this representation by $\kappa$, and one can easily show that: $$ \kappa(g)\cdot(x\cdot \psi)=(\lambda(g)\cdot x)\cdot(\kappa(g)\cdot \psi) $$ where $\lambda:\text{Spin}(n)\rightarrow \SO(n)$ is the double covering homomorphism.

My question is then this, does the above property uniquely determine the spinor representation up to isomorphism? I.e. if we have two faithful representations of $\text{Spin}(n)$ on $\mathbb{C}^N$ satisfying the property above, denoted by $\kappa$ and $\rho$, does there exist an isomorphism $F:\mathbb{C}^N\rightarrow \mathbb{C}^N$ satisfying: $$ \begin{align*} F(\kappa(g)\cdot \psi)=\rho(g)\cdot F(\psi) \end{align*} $$ for all $\psi\in \mathbb{C}^N$.

My motivation for this question is mostly due to the case where $n=4$. When $n=4$, we have that the space of Dirac spinors is given by $\mathbb{C}^4$, then we can obtain a representation of $\text{Spin}(4)$ on $\mathbb{C}^4$ as above. Indeed, if we choose an orthonormal basis for $\mathbb{R}^4$, it is not difficult to construct a faithful representation of $\text{Cl}(n)$ on $\mathbb{C}^4$, which then induces the isomorphism between the complex Clifford algebra. In particular, such a representation, when restricted to $\text{Spin}(n)$ has image in the subgroup of block diagonal matrices who's entries lie in $\SU(2)$. By construction, this representation satisfies the property above

However, we can also construct a representation of $\text{Spin}(4)$ in a different way, due to the exceptional isomorphism: $$ \begin{align*} \text{Spin}(4)\cong \SU(2)\times \SU(2) \end{align*} $$ We can then embed $\SU(2)\times \SU(2)$ in $\GL_4(\mathbb{C})$ by mapping $\SU(2)\times \SU(2)$ to block diagonal matrices, which naturally acts on $\mathbb{C}^4$. By identifying $\mathbb{R}^4$ with the real subspace of $2\times 2$ complex matrices spanned by the Pauli spin matrices, we can specify how $\mathbb{R}^4$ acts on $\mathbb{C}^4$. This action can be made to be the same as the one obtained by construction a faithful representation of $\text{Cl}(n)$ on $\mathbb{C}^4$. This representation of $\text{Spin}(4)$ on $\mathbb{C}^4$ then also respects Clifford multiplication with a vector.

Both of these representations are seemingly used interchangeably, though the latter is used more frequently when one wishes to discuss the specific case of $n=4$. Due to this, I suspect that the representations should be isomorphic, but I am unsure of how to show it.

If they are not isomorphic, then how do we justify using the latter representation when the former is derived in such generality?