Let $f : A_1 \to A_2$ be an injective homomorphism of unital Banach algebras. It's a standard fact that if $f$ is has closed range, i.e. $A_1$ is embedded as a closed subalgebra of $A_2$, then for every $x \in A_1$ we have $\rho(f(x)) = \rho(x)$, where $\rho$ denotes the spectral radius in the corresponding algebra.
Now, is the same true if $f$ does not have closed range - e.g. if its image is dense? I suspect that the answer is negative, but I failed to construct a counterexample...
A less precise but more intuitive formulation of this question would be: is the spectral radius - and, more generally, the polynomially convex hull of spectrum - an intrinsic property of operator, independent of the ambient algebra?
Please, see Banach subalgebra with a different spectrum Can this help you? In other words, consider polynomials on the interval $I=[1/2,3/4]$ with the uniform norm and with $\ell^1$-norm. The first norm leads to the $C^*$-algebra of continuous functions on $I$, the second norm leads to the Banach algebra of analytic functions inside the unite ball and restricted to the interval $I$. Hence $\rho_C(x)=3/4$ but $\rho_{\ell^1}(x)=1$ for the monomial $x$.