Can the Chebyshev polynomials evaluated at a zero for index m, form a sequence of fixed length (in the index) with even parity?

Question

For $n \in \mathbb{Z},n\geq 3$ find $x\in \mathbb{R}$ and index $m$ such that

1. $T_m(x)=0$,
2. $T_{m+j}(x)=T_{m-j}(x), \ j=1,\ldots, \lfloor \frac{n}{2} \rfloor -1$,
3. $T_{m+j}(x) = T_{m+n+1-j}(x), \ j=1,\ldots n.$

Note: condition 3. requires symmetry about $m+\frac{n+1}{2}$. However, I can find anti-symmetry, (insert a minus sign on rhs in 3.), for example, for $n=7$ and $m=2$, so $x=1/\sqrt{2}$ works.