FALSE: the left and right traces always agree

Let $\mathcal{C}$ be the tensor category of finite-dimensional $\mathbb{Z}$-graded vector spaces over a field $k$, and fix $q\in k^{\times }$ with $q\mathrm{\ne }\pm 1$. The usual double-dual map is a pivotal structure, and multiplying its component on degree-$n$ vectors by $q^n$ gives another pivotal structure $a^{\mathrm{\prime }}$.

Let $X=k[1]$ be the one-dimensional object concentrated in degree $1$. Then $X^{\vee }=k[-1]$, so ${\dim }_{a^{\mathrm{\prime }}}(X)=q,{\dim }_{a^{\mathrm{\prime }}}(X^{\vee })=q^{-1}\mathrm{.}$ Since $q\mathrm{\ne }{q}^{-1}$, the pivotal structure $a^{\mathrm{\prime }}$ is not spherical.

For the identity endomorphism $1_X$, the two trace expressions are exactly those two dimensions: ${\mathrm{Tr}}_L(a_X^{\mathrm{\prime }}{1}_X)=q,{\mathrm{Tr}}_R(1_X(a_X^{\mathrm{\prime }}{)}^{-1})=q^{-1}\mathrm{.}$ They are unequal, so left and right traces do not always agree. This also supplies a witness over $k=\mathbb{C}$, within the ground-field convention of In a spherical category the left and right traces agree; the pivotal structure is deliberately non-spherical.