I would like to prove that there is one and only one (non trivial) irreducible representation (up to equivalence) of the Clifford algebra $Cl(n)$ with $n=2p$ and $p\in\mathbb{Z}$. I have seen a proof which uses a sort of counting argument via the formulas $\sum_i|d_i|^2=|G|$ and also that the number of irreps is equal to the number of conjugacy classes. This proof is nice, but I would like to prove it in a different way using the following theorem:
Every automorphism of the associative algebra $\text{Mat}(t,\mathbb{C})$ is inner. Or in other symbols, if $h\in\text{Aut}(\text{Mat}(t,\mathbb{C}))$ then for all $M\in\text{Mat}(t,\mathbb{C})$, there exists some $S\in\text{GL}(t,\mathbb{C})$ such that$$h(M)=SMS^{-1}.$$
I will now try to prove the desired result. Let $T$ and $\widetilde{T}$ be two irreducible representations. I will denote the elements of the clifford algebra as $\{e_0,e_{i_1},e_{i_1}e_{i_2},\dots,e_{i_1}e_{i_2}\dots e_{i_{2p}} \}:=\{e_0,e_A\}$ and $\Gamma_i:=T(e_i)$, $\widetilde{\Gamma}_i:=T(e_i)$. I have already proven that the sets $\{\text{Id},\Gamma_A\}$ and $\{\text{Id},\widetilde{\Gamma}_A\}$ both form a basis for $\text{Mat}(2^p,\mathbb{C})$ where $\Gamma_A:=T(e_A)=T(e_{i_1}e_{i_2}\dots e_{i_r})=T(e_{i_1})T(e_{i_2})\dots T(e_{i_r})=$ $=\Gamma_{i_1}\Gamma_{i_2}\dots\Gamma_{i_r}$ for some $r\leq 2p$.
Define the following correspondence between the $\Gamma$ matrices, $\psi(\Gamma_i)=\widetilde{\Gamma}_i$. The correspondance $\psi$ is clearly bijective and induces an automorphism on $\text{Mat}(2^p,\mathbb{C})$, call it $\phi$, since under $\psi$, any $M:=\lambda_0\text{Id}+\sum_A\lambda_A\Gamma_A\in\text{Mat}(2^p,\mathbb{C})$ gets mapped to $M'=\phi(M)=\lambda_0\text{Id}+\sum_A\lambda_A\widetilde{\Gamma}_A$. Thus by the above theorem we conclude that there exists some $S\in\text{GL}(2^p,\mathbb{C})$ such that \begin{align} \phi(M)&=SMS^{-1},~~~~~~~~~~~~~~~~~~~~~~~~~~~~\forall ~M\in\text{Mat}(2^p,\mathbb{C})\\ \implies \lambda_0\text{Id}+\sum_A\lambda_A\widetilde{\Gamma}_A&=S\bigg(\lambda_0\text{Id}+\sum_A\lambda_A\Gamma_A\bigg)S^{-1}\\ \implies\sum_A\lambda_A\bigg(\widetilde{\Gamma}_A\bigg)&=\sum_A\lambda_A\bigg(S\Gamma_AS^{-1}\bigg). \end{align} From here I want to conclude that $\widetilde{\Gamma}_A=S\Gamma_AS^{-1}$ and hence the two irreps are actually equivalent, but I don't know what justifies me to conclude this (the coefficients $\lambda_A$ are worrying me). Have I taken the wrong approach if I want to prove this result via the above theorem?
Extra information: The theorem was taken out of ref [1] pg 432. Also, I am a physics student and haven't learnt about the ideals etc approach to Clifford algebras.
[1] B. Dubrovin, A Fomenko and S. Novikov, Modern Geometry Methods and Applications, Part I, Springer (1979).