I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S_{Y,Y} can be obtained from the symmetry S_{X,X}

as follows

$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.

Here, $c\circ c$ is shorthand for $(id_Y^{\otimes 2}\otimes c \otimes id)\circ(id_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.

In pictures, all I'm doing (which I would draw if I knew an easy way) is:

Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.

here is a pdf of the computation

I am just really repating a proof here that S_{U^,V^}=S_{U,V}^*, which holds for the braiding in any rigid braided monoidal categetory.

Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.

I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S_{Y,Y} can be obtained from the symmetry S_{X,X}

as follows

$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.

Here, $c\circ c$ is shorthand for $(id_Y^{\otimes 2}\otimes c \otimes id)\circ(id_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.

In pictures, all I'm doing (which I would draw if I knew an easy way) is:

Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.

here is a pdf of the computation

I am just really repating a proof here that $S_{U^*,V^*}=S_{U,V}^*$, which holds for the braiding in any rigid braided monoidal categetory.

Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.

I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S_{Y,Y} can be obtained from the symmetry S_{X,X}

as follows

$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.

Here, $c\circ c$ is shorthand for $(id_Y^{\otimes 2}\otimes c \otimes id)\circ(id_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.

In pictures, all I'm doing (which I would draw if I knew an easy way) is:

Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.

I am just really repating a proof here that S_{U^,V^}=S_{U,V}^*, which holds for the braiding in any rigid braided monoidal categetory.

Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.

I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S_{Y,Y} can be obtained from the symmetry S_{X,X}

as follows

$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.

Here, $c\circ c$ is shorthand for $(id_Y^{\otimes 2}\otimes c \otimes id)\circ(id_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.

In pictures, all I'm doing (which I would draw if I knew an easy way) is:

Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.

here is a pdf of the computation

I am just really repating a proof here that S_{U^,V^}=S_{U,V}^*, which holds for the braiding in any rigid braided monoidal categetory.

Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.