Jonathan wrote:

> I'm betting you're looking for \\(\mathcal{C}^\textbf{1} \cong \mathcal{C}\\).

Yes!

> It never ceases to amaze me how the same patterns always show up throughout mathematics.

Me neither! But category theory is designed to explain some of these patterns. Minutes ago, [over here](https://forum.azimuthproject.org/discussion/comment/19491/#Comment_19491), I confirmed Reuben's guess that \\(\mathbf{Cat}\\) is [cartesian closed](https://en.wikipedia.org/wiki/Cartesian_closed_category), meaning it has finite products and also exponentials. In any cartesian closed category we have

\[ x^1 \cong x \textrm{ and } 1^x \cong 1 \textrm{ and } (x \times y)^z \cong x^y \times x^z \textrm{ and } x^{y \times z} \cong (x^y)^z . \]

The most famous example is the category of sets; another is the category of finite sets, and that's ultimately the reason we learn these facts as equations in high school.

\\(\mathbf{Cat}\\) also has finite coproducts. In any cartesian closed category with finite coproducts we get other good things, like

\[ x^0 \cong 1 \textrm{ and } x^{y + z} \cong x^y \times x^z .\]

So, most of the isomorphisms you wrote down hold at this level of generality. However, it's not always true that \\(0^x \cong 0\\) for all \\(x \ncong 0\\).