Let $\cdots \to U\xrightarrow{g} V\xrightarrow{h} W\xrightarrow{i} X\xrightarrow{j}Y\to\cdots $ be an exact sequence of abelian groups and $W\cong W'$ for a certain abelian group $W'$. Can I then consider the sequence $\cdots \to U\xrightarrow{g} V\xrightarrow{h'} W'\xrightarrow{i'} X\xrightarrow{j}Y\to\cdots $ as an exact sequence as well?
(Or I would simply write $\cdots \to U\xrightarrow{g} V\xrightarrow{h'} W'\xrightarrow{i'} X\xrightarrow{j}Y\to\cdots $, somehow sloppy..)
I would guess yes, but I'm not 100% sure. Here with $i'$ and $h'$ I mean the following: If $\phi:W\to W'$ is the isomorphism, then $h'=\phi\circ h$ and $i'= i\circ \phi^{-1}$. It is $\operatorname{im}(h)=\ker(i)$ by assumption. I have to check that $\operatorname{im}(h')=\ker(i')$ and for this you need $\operatorname{im}(h')\cong \operatorname{im}(h)$ and $\ker(i)\cong \ker(i')$, i. e. several things are isomorphic, but not really equal...
Yes, it would still be an exact sequence. You mention that $\operatorname{im}(h')\cong\operatorname{im}(h)$ and $\ker(i)\cong \ker(i')$, but that these isomorphisms are not actually equalities; that's true, but these isomorphisms are actually induced by $\phi$ and its inverse, so in the end their composition is actually induced by the identity on $W'$, i.e. $\operatorname{im}(h')$ is really equal to $\ker(i')$.
To be more precise, you can prove the chain of equalities $$\operatorname{im}(h')=\phi(\operatorname{im}(h))=\phi(\ker(i))=\ker(i').$$ The first one is a consequence of the equality $h'=\phi\circ h$, the second one comes from the exactness of the first sequence and the last one from the equality $i'=i\circ \phi^{-1}$.