Please, help out of the mind trap. In this prominent paper Kadison and Ringrose prove among other things the following

Corollary 8. Each norm-continuous representation of a connected topological group by automorphisms of a $C^*$-algebra has range consisting of $\pi$-inner automorphisms.

By $\pi$-inner these authors mean permanently weakly inner, see the definition on page 4 of the same paper, for instance.

**Question:** Now suppose $A$ is a commutative $C^*$-algebra, and let $\pi:A\to\mathcal{B}(H)$ be a faithful $\star$-representation. A commutative $A$ has obviously no non-trivial inner automorphisms. The strong/weak closure $\pi(A)''\subset\mathcal{B}(H)$ is also commutative, so $\pi(A)$ cannot have even non-trivial weakly inner automorphisms. By the above corollary this would mean that $A$ has no norm-continuous group of automorphisms.

But let $A=C[0,1]$ for clarity (acting on $L^2[0,1]$), and let $y(x;t)=x^t$ for all $x\in[0,1]$ and $t\in(0,\infty)$. Now $y(.;t)$ is a homeomorphism of $[0,1]$ for every $t$, and therefore $\alpha_t(f)=y_*f=f[x^t]$ is definitely $\|f\|_{sup}$-preserving automorphism of $C[0,1]$, thus $\|\alpha_t\|_{\mathrm{op}}=1$. More generally, if $A\simeq C_0(M)$ by Gelfand-Naimark then every homeomorphism group $\phi_t\in\mathrm{Homeo}(M)$ gives such an automorphism group $\alpha_{\phi_t}$. In fact, the particular $\alpha_t$ above is not even spatial. The map $U:L^2[0,1]\to L^2[0,1]$ which gives $\alpha_t(f)=U_t f U_t^{-1}$ is also given by $U_t[h](x)=h(x^t)$ for all $h\in L^2[0,1]$, and is not unitary, because Lebesgue measure is not invariant under $y(.;t)$. $U_t$ is even unbounded for $t>1$.

So where does the source of confusion lie? Thank you.