There are three interesting types of upwards reflection. First off, reflecting upwards to an unbounded class of cardinals, or reflecting oneself upwards, or both. The other instance is when you can find a larger cardinal, but not an unbounded class. For instance, if $\kappa$ is $\gamma$-uplifting, then there is an inaccessible $\gamma'\gt\gamma$, and if $\kappa$ is superstrong, then there is some $\gamma'\gt\kappa$ such that $V_\kappa\prec V_{\gamma'}$.

If $\kappa$ is extendible or superhuge, and $\lambda$ is an ordinal then there is a proper class of $\gamma$ inaccessible, indescribable, measurable, Woodin, superstrong, and $\lambda$-supercompact. The reason for this is that, suppose $\kappa$ is inaccessible, and $\alpha$ is some ordinal. Then, if $\eta\gt\alpha$ and $j: V_{\kappa+\eta+1}\prec V_{j(\kappa+\eta)+1}$ with $j(\kappa)\gt\eta$, then $V_{\kappa+\eta+1}\vDash\kappa\text{ is inaccessible}$ and so $V_{j(\kappa+\eta)+1}\vDash j(\kappa)\text{ is inaccessible}$, and also $V_{j(\kappa)+1}\in V_{j(\kappa+\eta)+1}$, and inaccessibility is a $\Pi_1^1$ property of $V_{j(\kappa)}$.

Similarly Woodinness is a $\Pi_1^1$ property of $V_{j(\kappa)}$, and so if $\kappa$ is $1$-extendible then there is some Woodin cardinal above it. Measurability, on the other hand, is a $\Sigma_1^2$ property of $V_{j(\kappa)}$, and so if $\kappa$ is $2$-extendible, then there is a measurable cardinal above it, and full extendibility implies a proper class of measurable Woodin cardinals. Furthermore, $\Pi_n^m$ indescribability is a $\Sigma_n^m$ property of $V_{j(\kappa)}$, for $n\gt 0$ and $m\gt 1$, and so if $\kappa$ is $n$-extendible for $n\gt 1$, there is an $n$-indescribable, measurable, Woodin cardinal above $\kappa$.

Next, superstrongness is a bit trickier. Let $\alpha$ be some ordinal, $E$ an extender witnessing supestrongness, $\eta$ a limit $\gt\alpha+rank(E)$, and $j: V_{\kappa+\eta+1}\prec V_{j(\kappa+\eta)+1}$ with $j(\kappa)\gt\eta$. Then $V_{\kappa+\eta+1}\vDash\kappa\text{ is inaccessible}$ and so $V_{j(\kappa+\eta)+1}\vDash j(\kappa)\text{ is superstrong}$, and the proprety of being superstrong is absoloute between $V_\beta$ for $\beta$ a limit.

Finally, if $\kappa$ is $\lambda+1$-extendible, as witnessed by $j: V_{\kappa+\lambda+1}\prec V_{j(\kappa+\lambda)+1}$, then $j(\kappa)$ is $\lambda-$supercompact, because $P(P_\kappa(\lambda))\subseteq V_{j(\kappa+\lambda)+1}$, and so $D=\{X\subseteq P_\kappa(\lambda)|j"\lambda\in j(X)\}$ witnesses the $\lambda$-supercompactness of $j(\kappa)$.

This same argument can be applied to get that if $\kappa$ is extendible and huge (E.g. Superhuge), then there is a proper class of huge cardinals. Here is another way to get this. Let $P$ be some large cardinal witnessed by some structure of limited rank, i.e. $\Sigma_2$. Assume $P(\kappa)$ and $\kappa$ is extendible. Let $D$ be some witness to $P(\kappa)$, and $\eta$ a limit $\gt rank(D)$. Now, as $P(\kappa)\leftrightarrow\exists X(Q(\kappa,X))$, where $Q$ is $\Pi_1$, we have $V_{j(\kappa+\eta)}\vDash Q(\kappa,D)$ and so there is a normal measure on $\kappa$ concentrating on those cardinals which have $P(\lambda)$.

This a mirror of Reinhardt cardinals; if $\kappa$ is Reinhardt and $P(\kappa)$, then there is a normal measure concentrating on those cardinals which have $P(\lambda)$; if $\kappa$ is supercompact and $P(\kappa)$ is $\Sigma_2$, then there is a normal measure concentrating on those cardinals which have $P(\lambda)$. Furthermore, by if $j: V\prec M$ is a non-trivial elementary embedding with critical point $\kappa$, and $M^\kappa\subseteq M$, $P(\kappa)$ is $\Pi_1$, then there is a normal measure concentrating on those cardinals which have $P(\lambda)$ by downwards absoloutness of $\Pi_1$ formulas.

Now, if $\kappa$ is extendible, then it is $\Sigma_3$-reflecting. Then $V_\kappa\vDash\text{There is a proper class of }\lambda\text{ such that }P(\lambda)$. Therefore, as the previous assertion is $\Pi_3$, the same holds in $V$. Note that using extendibility and hugeness for a normal measure concentrating on the huge cardinals is a bit over shooting the goal; this follows from supercompactness and hugeness. Furthermore, if a cardinal is supercompact and rank-into-rank, there is a normal measure concentrating on those cardinals which are rank-into-rank, so that if $\kappa$ is extendible and rank-into-rank, then there is a proper class of rank-into-rank cardinals.

The other case is self-reflection. There are very few cases of self reflection. Typically, if a cardinal has some $\Pi_n$ property, the upwards reflection is derived from $\Sigma_n$-correction of $V_\kappa$ in $V$. This means that if $\kappa'\gt\kappa$ has that same property $P$, then $V_{\kappa'}\vDash P(\kappa)$. In fact, typically less is needed. For example, if $\kappa$ is extendible, then there is not necessarily a proper class of strongly unfoldable, strong, or supercompact cardinals (But there is a proper class of unfoldable cardinals, as every Ramsey cardinal is unfoldable).

However, in some instances this does not hold. Take rank-into-rank cardinals. If $j: V_\lambda\prec V_\lambda$ with critical point $\kappa$, then $j(\kappa)$ is also rank-into-rank, as witnessed by $j^+(j)$. In fact, each $j^n(\kappa)$ is rank-into-rank, as witnessed by $j^{n+}(j)$. Similar arguments work for $j: V_{\lambda+1}\prec V_{\lambda+1}$ and $j: L(V_{\lambda+1})\prec L(V_{\lambda+1})$, which are (Currently!) some of the strongest large cardinals. It is worth noting that the existence of some $j: L(V_{\lambda+1})\prec L(V_{\lambda+1})$ is in fact, $\Sigma_2$. It is equivalent to the existence of a normal non-principal $L(V_{\lambda+1})$-ultrafilter over $V_{\lambda+1}$, so that the least supercompact cardinals is larger than the least $I0$ cardinal, and if there is an extendible $I0$ cardinal, then there is a proper class of such cardinals.

Furthermore, if $j: V\prec V$ with critical point $\kappa$, then $j(\kappa)$ is also Reinhardt, as witnessed by $j^+(j)$. In fact, each $j^n(\kappa)$ is Reinhardt, as witnessed by $j^{n+}(j)$. However, it is not known if a Reinhardt cardinal need be supercompact or even $\Sigma_2$-reflecting, much less strongly extendible, though it must be weakly extendible. Furthermore, we can cut the universe off at the least inaccessible $\lambda\gt\kappa_\omega(j)$, to get $V_\lambda$ as a transitive model of $ZF$ with a Reinhardt cardinal.

Finally, we have large cardinals that self reflect upwards unboundedly many times. Two may knowledge, only one such large cardinal exist. If $\kappa$ is super Reinhardt with target $j(\kappa)$ above $\lambda$, and super Reinhardt with target $j'(\kappa)$ above $\alpha$, then $j(\kappa)$ is super Reinhardt with target above $\alpha$, as witnessed by $j^+(j')$. In fact, each $j^n(\kappa)$ is super Reinhardt with target above $\alpha$, as witnessed by $j^{n+}(j')$. Super Reinhardt cardinals are some of largest large cardinals. Every super Reinhardt cardinals is reflecting, superhuge, and a stationary limit of such cardinals. The existence of a super Reinhardt cardinal is also *significantly* stronger than that of an $I0$ cardinal.