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Here is a trial proof for the question over $Q_p$.

Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$).

Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers).

Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$.

Furthernote, $\bar\Phi_u\rightarrow\oplus J[\bar g(x)]$ gives m=up^v$for$\bar\Phi_{up^v}\rightarrow\oplus J[\bar g(x)^{\phi(p^v)}]$where (u,p)=1$. Further note, if $\phi(p^v)\neq 1$ as \bar\Phi_u=\prod \bar g$then$p\neq 2$. That \bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$, and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split, in other words this is the minimal polynomialof a block of . This follows since the reduction (mod $\bar\Phi_{up^v}$ corresponding to powers p$) of the companion matrix of$\bar g$f$ is as large as possible, namely itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$.

So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition , has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$.

From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[g(x^{\phi(p^v)})]$J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. This The general Jordan form classifies it the conjugacy type over a field, like as is$Q_p$. Note that,$\Phi_3\Phi_6$and$\Phi_6^2$give 4x4 matrices with order 6, failing for$p=2$. 2 added 93 characters in body Here is a proof for the question over$Q_p$. Write$J[f(x)^k]$for the general Jordan form of a irreducible$f$, being$k$identical blocks joined by 1's in general (minimal polynomial of block is$f^k$). Let$A$be finite order over$Q_p$, so$A\sim\oplus J[f(x)]$where the$f$have$f|\Phi_m$(cyclotomic polynomials) and finiteness implies the$f$are irreducible (not powers). Note$\bar f$determines$m$up to$p$-powers, writing$m=up^v$. Furthernote,$\bar\Phi_u\rightarrow\oplus J[\bar g(x)]$gives$\bar\Phi_{up^v}\rightarrow\oplus J[\bar g(x)^{\phi(p^v)}]$where$\phi(p^v)\neq 1$as$p\neq 2$. That is, the minimal polynomial of a block of$\bar\Phi_{up^v}$corresponding to powers of$\bar g$is as large as possible, namely$\bar g^{\phi(p^v)}$. So, every reduction to$\bar f$from the$A\sim\oplus J[f]$decomposition, has$\bar f(x)=\bar g(x)^{\phi(p^v)}$for some irreducible$\bar g|\bar\Phi_u$, that lifts to$g|\Phi_u$. From this,$\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$determines the general Jordan form of$A$uniquely as$A\sim\oplus J[g(x^{\phi(p^v)})]$. This classifies it over a field, like$Q_p$. Note that,$\Phi_3\Phi_6$and$\Phi_6^2$give 4x4 matrices with order 6, failing for$p=2$. 1 Here is a proof for the question over$Q_p$. Write$J[f(x)^k]$for the general Jordan form of a irreducible$f$, being$k$identical blocks joined by 1's in general (minimal polynomial of block is$f^k$). Let$A$be finite order over$Q_p$, so$A\sim\oplus J[f(x)]$where the$f$have$f|\Phi_m$(cyclotomic polynomials) and finiteness implies the$f$are irreducible (not powers). Note$\bar f$determines$m$up to$p$-powers, writing$m=up^v$. Furthernote,$\bar\Phi_u\rightarrow\oplus J[\bar g(x)]$gives$\bar\Phi_{up^v}\rightarrow\oplus J[\bar g(x)^{\phi(p^v)}]$where$\phi(p^v)\neq 1$as$p\neq 2$. That is, the minimal polynomial of a block of$\bar\Phi_{up^v}$corresponding to powers of$\bar g$is as large as possible, namely$\bar g^{\phi(p^v)}$. So, every reduction to$\bar f$from the$A\sim\oplus J[f]$decomposition, has$\bar f(x)=\bar g(x)^{\phi(p^v)}$for some irreducible$\bar g|\bar\Phi_u$, that lifts to$g|\Phi_u$. From this,$\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$determines the general Jordan form of$A$uniquely as$A\sim\oplus J[g(x^{\phi(p^v)})]$. This classifies it over a field, like$Q_p\$.