Are $\delta(x-5)$ and $\delta(5-x)$ the same function? As the title says, I'm confused whether $\delta(x-5)$ and $\delta(5-x)$ are the same function.

Similarly, are $\delta(x-5,y-6)$ and $\delta(5-x,6-y)$ the same function?

To, me it seems they are! But want a second opinion!
$\delta_t$ produces a distribution, meaning it generates a linear map $T_{\delta_t} : C_c^\infty \mapsto \mathbb{R}$, $\textit{defined}$ by

$$T_{\delta_t} (\phi) = &lt;\delta, \phi &gt; = \int \delta(t) \phi(t)dt :=\phi(0) $$
The first three equalities are all simply the same thing with different notation.The integral is more formal and makes computations more clear. In fact $\delta$ is homogeneous of degree -1 meaning that
$$\delta(\alpha x) = \frac{\delta(x)}{|\alpha|}$$
for $\alpha  \neq 0$. This can be shown by scaling the integral. So $\alpha = -1$ gives $\delta(x) = \delta(-x)$ as a distribution. Thus
$$T_{\delta_{x-5}}(\phi) = \int\delta(x-5)\phi(x)dx = \phi(5) = \int\delta(5-x)\phi(x) = T_{\delta_{5-x}}(\phi)$$
And similarly for the 2 dimensional case.
The expression $\delta(t)$ accepts as input a real number $t$, and returns as output either the number $1$, if $t=0$, or else the number zero, if $t\neq 0$. That is
$$
\delta(t)=\left\{
\begin{aligned}
1 &amp; \quad \text{if}\, t = 0 \\
0 &amp; \quad \text{if}\, t\neq 0
\end{aligned}
\right.
$$
If follows from this definition that $\delta(5-x)$ is equal to $1$ if and only if $5-x=0$, that is, if and only if $5=x$, and $\delta(x-5)$ is equal to $1$ if and only if $x-5=0$, that is, again, if and only $5=x$. Conclusion: $\delta(5-x)=\delta(x-5)$ for every real $x$, and more generally $\delta(a-b)=\delta(b-a)$ for every pair of real numbers $a,b$.