Left and right endomorphisms in monoidal categories

Let consider a monoidal category $\mathscr C, \otimes$ with unit $I$ and unitors $\lambda_X:I\otimes X\to X$ and $\varrho_X:X\otimes I\to X$.

Each endomorphism of the unit $a:I\to I$ induces, on each object $X$, the following endomorphisms: $$L(a,X)=\lambda_X^{-1}(a\otimes 1_X)\lambda_X:X\to X$$ $$R(a,X)=\varrho_X^{-1}(1_X\otimes a)\varrho_X:X\to X$$

Then both $L(a,X)$ and $R(a,X)$ are natural in $X$, and $L(a,I)=a=R(a,I):I\to I$. Note that such equality holds for all objects if $I$ is a generator for the category, that's $L(a,X)=R(a,X)$. Indeed for each $x:I\to X$, by naturality, we have: $L(a,X)\circ x=x\circ a=R(a,X)\circ x$.

Question: the equality $L(a,X)=R(a,X)$ holds for all objects $X$ (even without assuming $I$ to be a generator for the category)?

By no means. Think of the category of bimodules over a commutative ring $k$, so $k$-vector spaces with two actions which are associative with respect to each other but which may not agree. For instance $\mathbb{R}[\epsilon]$ has a natural extra action on all its modules by quotients go down to $\mathbb{R}$ and using the vector space action. Then $L(a,X)$ corresponds to multiplying on the left, and $R(a,X)$ on the right, with an element of $k$, which in the example, acting on itself, is the difference for instance between $x+y\epsilon\mapsto 0,\mapsto x\epsilon$.