I am trying to show that if $\sim^*$ denotes equivalence in the categorical sense, then; $$C \sim^* D \sim^* E \implies C\sim^* E$$
For functors we will use $\sim$ to denote a natural isomorphism.
So we have functors$~~$ $T:C \rightarrow D$,$~~~$ $S:D \rightarrow C$ s.t. $TS \sim 1_D$ and $ST \sim 1_C$, and also
$~~$ $L:D \rightarrow E$,$~~~$ $K:E \rightarrow D$ s.t. $LK \sim 1_E$ and $KL \sim 1_D$.
I now want to show that $SKLT \sim 1_C$ and $LTSK \sim 1_E$.
Just focusing on the first of these we have that on objects $SKLT(u) \cong ST(u) \cong u$ $(\dagger)$where the first isomorphism comes from $KL \sim 1_E$ applied to $T(u)$. So we now only need to show that the relevant diagrams commute, i.e. if $f: u\rightarrow v$ then $$f \alpha _u= \alpha _v SKLT(f),$$
where $\alpha _*$ are the isomorphisms between the objects we showed exist at $(\dagger)$. This is the bit that I have become stuck at and I would appreciate any help and hints that people can give me.
I have managed to solve my own question (I think) so I'll post what I did here to hopefully help others searching for the same thing in future.
So we know that $KL \sim 1_D$ $(\star)$, for objects $u$ and $v$ in $C$ we consider $T(u)$ and $T(v)$, two objects in $D$. If we also have $f:u \rightarrow v$ then $Tf:T(U) \rightarrow T(V)$ (a morphism in D). So by $(\star)$ $\exists$ $\alpha_{T(u)}$ and $\alpha_{T(v)}$ isomorphisms $KL(T(u)) \rightarrow T(u)$ and $KL(T(v)) \rightarrow T(v)$ respectively s.t. $$ T(f) \alpha_{T(u)} = \alpha_{T(v)}KL(T(f)) $$
Then if we apply the functor $S$ everywhere around the diagram that holds the above information (i.e. to maps and objects) and observe that a functor preserves isomorphisms, we now have a natural isomorphism between $SKLT$ and $ST$. Again imagining this information in the commutative diagram form we can "glue" the commutative diagram for $ST$ being naturally isomorphic to the identity on to this diagram at the right hand side. Then this whole diagram must commute (easy to show) and hence it in fact gives a natural isomorphism between $SKLT$ and $1_C$ (since composition of isomorphisms is an isomorphism). The proof for $LTSK \sim 1_E$ is then similar.