Transfer rank2 tensor to a new basis

The trace of the sigma should be the same in both new and old basis. But I get a different one. Really appreciate for the help.
I’ll put the screen shot in the comment part

I have no idea about your calculations since I haven't really learned the rules of manipulations of tensors. However, I am surprised that you don't get off-diagonal entries although the new basis looks as if there should be several of them.

I see that
$$
\sigma_{ij}=\underline{\hat{e}}^{(1)} \otimes \underline{\hat{e}}^{(1)}+\underline{\hat{e}}^{(2)}\otimes \underline{\hat{e}}^{(2)}+2\cdot\underline{\hat{e}}^{(3)}\otimes \underline{\hat{e}}^{(3)}
$$
Hence, I would write ##\underline{\hat{e}}^{(k)}=\alpha_k \underline{\hat{e}}'^{(1)}+\beta_k \underline{\hat{e}}'^{(2)}+\gamma_k \underline{\hat{e}}'^{(3)},## determine the ## \alpha_k\, , \,\beta_k\, , \,\gamma _k,## substitute all of them in the first equation and rearrange everything with the distributive law to obtain an equation
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$