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I answer the question in the title: $T$ can act trivial iff the rep is trivial.

Here is the proof:

Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') = Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.

Determining the eigenspaces, you need to compute $$ Res_N \sigma.$$ This is easy with Mackey induction restriction formula with expensive bookkeeping for those induced from $B$. For the supercuspidals, I think things become much worse. I also suggest working with GL(2) and study Bushnell-Kutzko if you can. I don't see an easier way. Usually $N$ squarefree is easier, because you don't need primes powers, and can work with a finite field.

I answer the question in the title: $T$ can act trivial iff the rep is trivial.

Here is the proof:

We have $$ SL_2(Z/N) = \oplus_{p^k || N} SL(Z/p^k)$$ by the Chinese remainder theorem and because SL(2) is an algebraic group. So you have to work with $G=SL_2(Z/p^k)$ only.

Have understood this: Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') = Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.

Determining the eigenspaces, you need to compute $$ Res_N \sigma.$$ This is easy with Mackey induction restriction formula with expensive bookkeeping for those induced from $B$. For the supercuspidals, I think things become much worse. I also suggest working with GL(2) and study Bushnell-Kutzko if you can. I don't see an easier way. Usually $N$ squarefree is easier, because you don't need primes powers, and can work with a finite field.

Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') = Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.

I answer the question in the title: $T$ can act trivial iff the rep is trivial.

Here is the proof:

Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') = Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.

Determining the eigenspaces, you need to compute $$ Res_N \sigma.$$ This is easy with Mackey induction restriction formula with expensive bookkeeping for those induced from $B$. For the supercuspidals, I think things become much worse. I also suggest working with GL(2) and study Bushnell-Kutzko if you can. I don't see an easier way. Usually $N$ squarefree is easier, because you don't need primes powers, and can work with a finite field.

Assume wlog that the rep $\sigma$ and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom(1,Res_{N} \sigma') \neq C^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.

Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') = Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial.