Let $(M,\otimes,1)$ be closed monoidal category and $C$ an $M$-enriched category. Assume we have $C$-objects $X$ and $X'$ and a morphism $f:1\to C(X,X')$ in $M$. We call $f$ an *isomorphism* if there is a $g:1\to C(X',X)$ in $M$ such that
$$1\cong 1\otimes 1\stackrel{g\otimes f}{\to} C(X',X)\otimes C(X,X')\to C(X,X)$$
equals $\mathrm{id}:1\to C(X,X)$ and the same for $C(X',X')$. Now for each $Y$, $f$ induces a morphism
$$f^*_Y:C(X',Y)\cong C(X',Y)\otimes 1\stackrel{\mathrm{id}\otimes f}{\to} C(X',Y)\otimes C(X,X')\to C(X,Y).$$
Assume that $f_Y^*$ is an $M$-isomorphism for all $Y$. As in the basic case ($C$ enriched over $\mathbf{Set}$), I want to conclude that $f$ is an isomorphism. To construct an inverse $g$, we should use
$$g:1\stackrel{\mathrm{id}}{\to} C(X,X)\stackrel{(f_X^*)^{-1}}{\to} C(X',X).$$

First of all: Is this correct? At some point, it seems that I have to use the (enriched) naturality of $f^*$, now as a map $f^*_Y:1\to C(X,Y)^{C(X',Y)}$.