let $A , B $ be a banach algebra with identity, and $ \varphi : A \longrightarrow B $ is nonzero homoemorphism so that $ \varphi ( 1_{A} ) = 1 _{B}$
can we say?
a: $ \sigma( \varphi ( a )) \subseteq \sigma ( a ) $
b: $ \varphi ( G ( A)) \subseteq G ( B )) $
$( G (A) = \{ a \in A : \textrm{a is invertible} \} ) $
We can certainly state that b holds. In particular: if $a \in A$ has inverse $a^{-1}$, then $\varphi(a)\varphi(a^{-1}) = \varphi(a^{-1}) \varphi(a) = 1_B$, which means that $\varphi(a) \in G(B)$.
Spectrum is a little trickier. We have $$ \lambda \notin \sigma(a) \implies\\ (a - \lambda I) \in G(A) \implies\\ \varphi(a - \lambda I) \in G(B) \implies\\ \varphi(a) - \lambda I \in G(B) \implies\\ \lambda \notin \sigma(\varphi(a)) $$ Thus, by contrapositive, we have $\lambda \in \sigma(\varphi(a)) \implies \lambda \in \sigma(a)$. That is, $\sigma(\varphi(a)) \subseteq \sigma(a)$, as was desired.