$\DeclareMathOperator\op{op}$Let $A$ be a $C^\*$-algebra. We know that $A$ admits a natural operator space structure, namely the operator space structure induced by any faithful $\*$-representation of $A$ into some $B(\mathcal H)$ for some Hilbert space $\mathcal H.$ Now consider the opposite $C^\*$ algebra $A^{\op}$ which is the same as $A$ but multiplication reversed. Now $A^{\op}$ has a natural operator space structure as well.


1. Is it true that the matrix norms are given by $\|[x\_{ij}]\_{i,j=1}^n\|\_{M\_n(A^{\op})}=\|[x\_{ji}]\_{i,j=1}^n\|\_{M\_n(A)}$?
2. Suppose $A$ and $A^{\op}$ is isomorphic as $C^\*$-algebras then there exists some constant $C>0$ such that for all $n\geq 1$ and $[x\_{ij}]\_{i,j=1}^n\in M\_n(A)$ we have $\|[x\_{ij}]\_{i,j=1}^n\|\_{M\_n(A)}\leq C \|[x\_{ji}]\_{i,j=1}^n\|\_{M\_n(A)}$?



Opposite $C^*$ algebras
1. Yes. You could consult [Pisier's Operator Space Theory book](https://www.cambridge.org/core/books/introduction-to-operator-space-theory/DE174FA28C7DBC243FBEA3911E97EA4E) sections 2.9 and 2.10.


To give details: let $H$ be a Hibert space and $\overline H$ be the complex conjugate space. This has vectors $\{ \overline\xi : \xi\in H \}$ with the same addition, the complex conjugate multiplication, and the inner product $(\overline\xi|\overline\eta)\_{\overline H} = (\eta|\xi)\_H$. For $x\in\mathcal{B}(H)$ define $x^\top\in\mathcal{B}(\overline H)$ by $x^\top{\overline\xi} = \overline{x^\*(\xi)}$. Then $x\mapsto x^\top$ is an anti-homomoprhism, and we have a coordinate-free notion of the transpose.


Given a (faithful) $\*$-representation $\pi:A\rightarrow\mathcal B(H)$ define $\pi^\top:A^{op}\rightarrow\mathcal B(\overline H)$ by $\pi^\top(x) = \pi(x)^\top$. Then $\pi^\top$ is a faithful) $\*$-representation of $A^{op}$. Let $\pi$ be faithful. Given $x = (x\_{ij})\in M\_n(A^{op})$ and $\xi=(\overline{\xi\_j})$ and $\eta=(\overline{\eta\_i})$ in $\overline{H}^n$,
$$ (\eta|(\operatorname{id}\otimes\pi^\top)(x)\xi)\_{\overline{H}^n}
= \sum\_{i,j} (\overline{\eta\_i}|\pi(x\_{ij})^\top \overline{\xi\_j} )\_{\overline{H}}
= \sum\_{i,j} (\overline{\eta\_i}|\overline{\pi(x\_{ij}^\*)\xi\_j} )\_{\overline{H}}
= \sum\_{i,j} (\pi(x\_{ij}^\*)\xi\_j | \eta\_i )\_H
= \sum\_{i,j} (\xi\_j | \pi(x\_{ij})\eta\_i )\_H. $$
As $\operatorname{id}\otimes\pi^\top$ induces the norn on $M\_n(A^{op})$, and $\operatorname{id}\otimes\pi$ induces the norn on $M\_n(A)$, it follows that $\|x\|\_{M\_n(A^{op})} = \| (x\_{ji}) \|\_{M\_n(A)}$.


2. I don't think so. Fix $n$, and consider the matrix units $e\_{ij}\in M\_n$. Consider $x = \sum\_{i,j} e\_{ij} \otimes e\_{ji} \in M\_n\otimes M\_n$. Being careful, if $A=M\_n$, then $x = (x\_{ij}) \in M\_n(A)$ has $x\_{ij} = e\_{ji}$. Thus $(x\_{ji}) = \sum\_{i,j} e\_{ij} \otimes e\_{ij} \in M\_n\otimes M\_n$. Using that $M\_n\otimes M\_n \cong \mathcal{B}(C^{n^2})$, for $\xi = \xi\_{a,b} \in \mathbb C^n\otimes \mathbb C^n \cong \mathbb C^{n^2}$ we see that
$$ x(\xi) = \sum\_{i,j,a,b} \xi\_{a,b} e\_{ij}(e\_a) \otimes e\_{ji}(e\_b)
= \sum\_{i,j} \xi\_{j,i} e\_i \otimes e\_j
\quad\implies \|x(\xi)\| = \|\xi\|, $$
while
$$ (x\_{ji})(\xi) = \sum\_{i,j,a,b} \xi\_{a,b} e\_{ij}(e\_a) \otimes e\_{ij}(e\_b)
= \sum\_{i,j} \xi\_{j,j} e\_i \otimes e\_i, $$
from which it follows that $\|x\| = n$.


However, $M\_n$ and $M\_n^\op$ are isomorphic via the transpose map $\theta(x) = x^t$. Check: $\theta$ is linear, a homomorphism, and $\theta(x^\*) = (x^\*)^t = \overline{x} = (x^t)^\* = \theta(x)^\*$.


So a counter-example comes from letting $A$ be the $c\_0$-direct sum $\bigoplus\_n M\_n$.


