Injective envelopes are brutal.

Let $A=\text{UHF}(2^\infty)$ and $B$ the hyperfinite II$_1$ factor. Take $f$ to be the inclusion map. We have $I(B)=B$, while $I(A)$ is a wild AW$^*$ factor of type III.

If $g:I(A)\to B$ is a $*$-homomorphism and $\tau$ is the trace on $B$, then $\gamma=\tau\circ g$ is a trace on $I(A)$. In a type III AW$^*$-factor any projection $p$ can be halved, so there exist $p_1,p_2$ with $p=p_1+p_2$ and $p\sim p_1\sim p_2$, which gives us the usual
$$ \gamma(p)=\gamma(p_1)+\gamma(p_2)=2\gamma(p), $$ and so $\gamma(p)=0$ for any projection $p$. Thus $\gamma=0$. As $\tau$ is faithful, $g=0$.

In summary, $f$ is a $*$-monomorphism that admits no extension to a $*$-homomorphism, and in fact the only $*$-homomorphism $I(A)\to I(B)$ is the zero homomorphism.

As mentioned by Chris, injective envelopes are brutal when seen as C$^*$-algebras.

Let $A=\text{UHF}(2^\infty)$ and $B$ the hyperfinite II$_1$ factor. Take $f$ to be the inclusion map. We have $I(B)=B$, while $I(A)$ is a wild AW$^*$ factor of type III.

If $g:I(A)\to B$ is a $*$-homomorphism and $\tau$ is the trace on $B$, then $\gamma=\tau\circ g$ is a trace on $I(A)$. In a type III AW$^*$-factor any projection $p$ can be halved, so there exist $p_1,p_2$ with $p=p_1+p_2$ and $p\sim p_1\sim p_2$, which gives us the usual
$$ \gamma(p)=\gamma(p_1)+\gamma(p_2)=2\gamma(p), $$ and so $\gamma(p)=0$ for any projection $p$. Thus $\gamma=0$. As $\tau$ is faithful, $g=0$.

In summary, $f$ is a $*$-monomorphism that admits no extension to a $*$-homomorphism, and in fact the only $*$-homomorphism $I(A)\to I(B)$ is the zero homomorphism.