# A question about limit cones and isomorphism

A consider a limit cone on a diagram $$D: \mathbf{I} \rightarrow \mathbf{C}$$:

$$\left( L \xrightarrow{p_i} D(I) \right)_{I \in \mathbf{I}}$$

Now suppose that $$L' \in \mathbf{C}$$ is some object isomorphic to $$L$$, and that there is some cone:

$$\left( L' \xrightarrow{p_i'} D(I) \right)_{I \in \mathbf{I}}$$

Is it necessarily the case that this cone is also a limit cone?

It's easy to see that any isomorphism from $$L \xrightarrow{\phi} L'$$ gives rise to a limit cone with vertex $$L'$$ simply by writing:

$$\left( L' \xrightarrow{p_i \circ \phi^{-1}} D(I) \right)_{I \in \mathbf{I}}$$

But it seems not at all necessary that this cone and the previous one are the same and/or ismorphic as cones. Yet, it seems as if any cone whose vertex is isomorphic to the limit should be a limit cone; that is, that the uniqueness of limit cones should "extend up" to uniqueness of limit objects.

No, this doesn't follow at all. For example, consider $$L=\mathbb N\times\mathbb N$$, i.e. the product of $$\mathbb N$$ and $$\mathbb N$$ (in $$\mathbf{Set}$$). A cone for the corresponding diagram is just two arrows $$L'\to\mathbb N$$. Let's pick $$L'=\mathbb N\times\mathbb N$$ which is surely isomorphic to $$L$$. Now, let's pick the two arrows of the cone out of $$L'$$ to both be $$n\mapsto 0$$. This is obviously not a limit cone.
What you would want is not an isomorphism of the apex of the cones, but an isomorphism of the cones (in a suitable category of cones). A morphism of cones is a morphism of the apex objects, $$\varphi:L\to L'$$, which satisfies the condition $$p'_i \circ \varphi = p_i$$ for every $$i\in\mathbf{I}$$. Clearly an isomorphism of cones will lead to exactly the cones induced by the isomorphism of apex objects you describe in the latter part of your question. If you have such an isomorphism of cones, they will, as you state, both be limit cones if one is.