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\$$.